Vandermonde行列式的平方导出的恒等式

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289. Nice identities in an algebra/ geometry/trigonometry.

by Virgil Nicula

PP1. Suppose that $\{a,b,c\}\subset\mathbb C$ such that $a+b+c=0$ . Prove that $2(a-b)^2(b-c)^2(c-a)^2=(a^2+b^2+c^2)^3-54a^2b^2c^2$ .

Proof. Are well-known the Viete's relations $t^3-s_1t^2+s_2t-s_3=0\begin{array}{c} \nearrow\\\ \rightarrow\\\ \searrow\end{array}\begin{array}{c} a\\\\ b\\\\ c\end{array}\ \iff\ \left\|\begin{array}{c} s_1=a+b+c\\\\ s_2=ab+bc+ca\\\\ s_3=abc\end{array}\right\|$ . Denote $S_k=a^k+b^k+c^k$ , where $k\in\mathbb N$ . Observe that $E\equiv (a-b)^2(b-c)^2(c-a)^2=$ $\left|\begin{array}{ccc} 1 & 1 & 1\\\ a & b & c\\\ a^2 & b^2 & c^2\end{array}\right|^2=$ $\left|\begin{array}{ccc} 1 & 1 & 1\\\ a & b & c\\\ a^2 & b^2 & c^2\end{array}\right|\cdot\left|\begin{array}{ccc} 1 & a & a^2\\\ 1 & b & b^2\\\ 1 & c & c^2\end{array}\right|=$ $\left|\begin{array}{ccc} S_0 & S_1 & S_2\\\ S_1 & S_2 & S_3\\\ S_2 & S_3 & S_4\end{array}\right|=$ $\left|\begin{array}{ccc} 3 & 0 & S_2\\\ 0 & S_2 & S_3\\\ S_2 & S_3 & S_4\end{array}\right|=$ $3S_2S_4-S_2^3-3S_3^2$ . Since $s_1=0\implies \left\|\begin{array}{ccc} S_2 & = & -2s_2\\\\ S_3 & = & 3s_3\\\\ S_4 & = & 2s_2^2\end{array}\right\|$ obtain that

$E=3\left(-2s_2\right)\left(2s_2^2\right)-\left(-2s_2\right)^3-3\left(3s_3\right)^2=$ $-4s_2^3-27s_3^2$ . In conclusion, $2(a-b)^2(b-c)^2(c-a)^2=2E=$ $-8s_2^3-54s_3^2=$ $S_2^3-54s_3^2=$ $\left(a^2+b^2+c^2\right)^3-54a^2b^2c^2$ .

Particular case. For any $x\in\mathbb R$ exists the inequality $(x+1)^2+x^2+1\ \ge\ 3\cdot\sqrt [3]{2x^2(x+1)^2}$ .


PP2. Prove that in a $ ABC$ triangle we have the relation $ \boxed{a\tan\frac A2+b\tan\frac B2+c\tan\frac C2=2(2R-r)}$ .

Proof 1. $\sum a\cdot\tan\frac A2=\sum\frac {ar}{s-a}=$ $\frac {r}{\prod (s-a)}\cdot \sum a(s-b)(s-c)=$ $\frac {r}{sr^2}\cdot 2sr(2R-r)=2(2R-r)$ .

Remark. I used the well-known identity $\boxed{\sum a(s-b)(s-c)=2sr(2R-r)}$ which is equivalently with the identity $\boxed{\sum ar_a=2s(2R-r)}\ (*)$ .

Indeed, from the relation $S=s\cdot r=(s-a)\cdot r_a$ obtain that $ar_a=s(r_a-r)$ . I supposed well-known that $\sum r_a=4R+r$ .

Proof 2. $\sum a\tan\frac{A}{2}=$ $2R\cdot\sum 2\sin^2\frac A2=$ $2R\cdot\sum \left(1-\cos A\right)=$ $2R\cdot\sum \left(3-\sum\cos A\right)=$ $2R\left(2-\frac rR\right)=2(2R-r)$ .

Remark. I used the well-known identity $\boxed{\sum\cos A=1+\frac rR}$ . Indeed, $\sum\cos A=$ $1-2\sin^2\frac A2+2\sin\frac A2\cos\frac {B-C}{2}=$

$1+2\sin\frac A2\left(\cos \frac {B-C}{2}-\cos\frac {B+C}{2}\right)=$ $1+4\prod\sin\frac A2=$ $1+4\prod\sqrt {\frac {(s-b)(s-c)}{bc}}=$ $1+\frac {\prod (s-a)}{abc}=$ $1+\frac {sr^2}{4Rsr}=1+\frac rR$ .

Proof 3. $ \left\|\begin{array}{cc} r_a = s\cdot\tan\frac A2 & (1) \\ \\ \ ar_a = s\left(r_a - r\right) & (2) \\ \\ r_a + r_b + r_c = 4R + r & (3)\end{array}\right\|\implies$ $ \sum a\cdot\tan\frac A2\ \stackrel {(1)}{ = }\ \sum \frac {ar_a}{s}\ \stackrel{(2)}{ = }\ \sum\left(r_a - r\right)\ \stackrel{(3)}{ = }\ 2(2R - r)$ .

Proof 4. $ \left\|\begin{array}{ccc} S = s(s - a)\tan\frac A2 & (4) \\ \\ S = sr = (s - a)r_a & (5)\end{array}\right\|\implies$ $ \sum a\cdot\tan\frac A2\ \stackrel{(4)}{ = }\ S\cdot\sum\frac {a}{s(s - a)} =$ $ \sum\left(\frac {S}{s - a} - \frac {S}{s}\right)\ \stackrel{(5)}{ = }\ \sum\left(r_a - r\right) = 2(2R - r)$ .

Remark. $ \sum\frac {a}{r_a}\ \ge\ \frac {2s}{2R - r}$ . Define $ S_n = \sum a^n\cdot\tan\frac A2\ ,\ n\in \mathrm N$ . Then $ \boxed {\ S_{n + 1} = s\cdot S_n - r\cdot\sum a^n\ ,\ n\in \mathrm N\ }$ .

Indeed, the first relation becomes $\sum\frac a{r_a}=\sum\frac {a^2}{ar_a}\stackrel{(\mathrm{C.B.S.})}{\ge}\frac {\left(\sum a\right)^2}{\sum ar_a}\stackrel{(*)}{=}\frac {4s^2}{2s(2R-r)}=\frac {2s}{2R-r}$ and the second relation becomes

$S_{n+1}=\sum a^{n+1}\tan\frac A2=$ $\sum a^n\cdot\frac {ar_a}{s}=$ $\sum a^n(r_a-r)=$ $\sum a^nr_a-r\sum a^n=$ $s\cdot\sum a^n\tan\frac A2-r\cdot\sum a^n=$ $s\cdot S_n-r\cdot\sum a^n$ .


PP3. Ascertain the monic polynomial equation which has the roots $\cos \frac{\pi}{7}\ ,\ \cos \frac{3\pi}{7}\ ,\ \cos \frac{5\pi}{7}$ .

Proof. Using the remarkable sum $\boxed{ x_{k+1}=x_k+r\ ,\ k\in\overline {1,n}\implies \sum_{k=1}^n \cos x_k=\frac {\cos\frac {x_1+x_n}2\sin\frac {nr}2} {\sin\frac r2} }$ obtain that $\boxed{s_1\equiv\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}}=\frac {\cos\frac {3\pi}{7}\sin\frac {3\pi}{7}}{\sin\frac {\pi}{7}}=\frac {\sin\frac {6\pi}{7}}{2\sin\frac {\pi}{7}}=\frac {\sin\frac {\pi}{7}}{2\sin\frac {\pi}{7}}\implies$

$\boxed{\ s_1\ =\ \frac 12\ }\ (*)$ . Otherwise. $2\sin\frac {\pi}{7}\cdot s_1=$ $2\sin\frac {\pi}{7}\cdot\left(\cos \frac{\pi}{7}+\cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}\right)=$ $\sin\frac {2\pi}{7}+\sin\frac {4\pi}{7}-\sin\frac {2\pi}{7}+\sin\frac {6\pi}{7}-\sin\frac {4\pi}{7}=\sin\frac {6\pi}{7}=\sin\frac {\pi}{7}\implies$ $s_1=\frac 12$ .

$\boxed{s_2\equiv\cos \frac{\pi}{7}\cdot\cos \frac{3\pi}{7}+\cos \frac{3\pi}{7}\cdot\cos \frac{5\pi}{7}+\cos \frac{5\pi}{7}\cdot \cos \frac{\pi}{7}}\implies$ $2s_2=\cos\frac {4\pi}{7}+\cos\frac {2\pi}{7}+$ $\cos\frac {8\pi}{7}+\cos\frac {2\pi}{7}+$ $\cos\frac {6\pi}{7}+\cos\frac {4\pi}{7}=$ $-\cos\frac {3\pi}{7}+\cos\frac {2\pi}{7}-\cos\frac {\pi}{7}+\cos\frac {2\pi}{7}-\cos\frac {\pi}{7}-\cos\frac {3\pi}{7}=$

$-2\left[\left(\cos\frac {3\pi}{7}+\cos\frac {\pi}{7}\right)-\cos\frac {2\pi}{7}\right]\stackrel{(*)}{=}$ $-2\left[\left(\frac 12-\cos\frac {5\pi}{7}\right)+\cos\frac {2\pi}{7}\right]=$ $-1+2\left(\cos\frac {5\pi}{7}+\cos\frac {2\pi}{7}\right)\implies$ $\boxed{\ s_2\ =\ -\frac 12\ }$ . $\boxed{s_3\equiv\cos \frac{\pi}{7}\cdot\cos \frac{3\pi}{7}\cdot\cos \frac{5\pi}{7}}\implies$

$4s_3=2\cos\frac {5\pi}{7}\left(\cos\frac {4\pi}{7}+\cos\frac {2\pi}{7}\right)=$ $\cos\frac {9\pi}{7}+\cos\frac {\pi}{7}+\cos\frac {7\pi}{7}+\cos\frac {3\pi}{7}=$ $-\cos\frac {2\pi}{7}-1+\left(\cos\frac {\pi}{7}+\cos\frac {3\pi}{7}\right)\stackrel{(*)}{=}$ $-\cos\frac {2\pi}{7}-1+\left(\frac 12-\cos\frac {5\pi}{7}\right)=-\frac 12\implies$ $\boxed{\ s_3\ =\ -\frac 18\ }$ .

In conclusion, $\cos \frac{\pi}{7}\ ,\ \cos \frac{3\pi}{7}\ ,\ \cos \frac{5\pi}{7}$ are the roots of the equation $x^3-\frac 12x^2-\frac 12x+\frac 18=0$ , i.e. $8x^3-4x^2-4x+1=0$ .


PP4. Let $ ABCD$ be a convex quadrilater with $ \angle DAC=10^{\circ},\ \angle DCA=20^{\circ},\ \angle BAC=30^{\circ}$ and $ \angle BCA=50^{\circ}$. Prove that $ AC\perp BD$ .

Proof 1 ("slicing"). . I"ll use the well-known lemma "Let $ABCD$ be a convex quadrilateral so that $\left\|\begin{array}{c} DA=DC\\\ D+2\cdot B=360^{\circ}\end{array}\right\|$ . Then $DA=DB$" (you can prove easily).

Denote $E\in (AD)$ so that $\widehat{ECA}\equiv\widehat{ECD}$ . Since $EA=EC$ and $m\left(\widehat{AEC}\right)+2\cdot m\left(\widehat{ABC}\right)=$ $160^{\circ}+2\cdot 100^{\circ}=360^{\circ}$ obtain from upper lemma

that $EB=EA$ , i.e. $EB=EA=EC=BC$ . Since $BC=BE$ and $m\left(\widehat{EBC}\right)+2\cdot m\left(\widehat{EDC}\right)=$ $60^{\circ}+2\cdot 150^{\circ}=360^{\circ}$ obtain again from

upper lemma that $BD=BC$ , i.e. $EB=EA=EC=BC=BD$ . In conclusion, $BD=BC$ and $m\left(\widehat{CBD}\right)=40^{\circ}$ , i.e. $BD\perp AC$ .

Proof 2 ("slicing"). Let's draw equilateral triangle $ACY$ ,where the poiny $Y$ lies on the same semi-plane with the point $B$w.r.t. $AC$ . Hence $ m(\angle ACB)=50^\circ => $

$ m(\angle BCY)=10 ^\circ $ . Hence $ \angle YAB=30^\circ =\angle BAC =>$ the streight line $AB$ is the symmetric line of side $YC$ . Then $ \angle BYC=\angle BCY=10^\circ$ . Let

$ m(\angle XCA)=m(\angle XAC)=10^\circ (X\in AD)$ . Hence $ AC=YC =>$ $ \Delta XAC \equiv \Delta YBC$ $ =>$ $ XA=XC=YB=BC$ . The symmetric

line of the side $AC$ will pass throught the points $X$ and $Y$ . $ => $ $ m(\angle XYA)=30^\circ$ and hence $m(\angle BYC)=10^\circ => $ $m\angle XYB)=20^\circ$ . Let's draw

a line dividing $ \angle XYB$ into two equal parts and let's this line crosses the straight line $AD$ in point $W$ . Hence $m(\angle XYB)=20^\circ =>$ $m(\angle AYW)=40^\circ$ .Then

$ \Delta AYW$ will be an isosceles triangle $ m(\angle WAY)=m(\angle AWY)=70 20^\circ )$ . $ => $ $AY=YW=CY=>$ and the triangle $WYC$ will be an isosceles triangle.
$m(\angle WYC)=20 ^\circ =>$ $ m(\angle YWC)=m(\angle YCW)=80^\circ$ , but at the same time and $m(\angle YCD)=80^\circ =>$ $ D \equiv W$ . Then $m(\angle ADY)=70 ^\circ$ and

$ \Delta DBY \equiv \Delta CBY => $ $ m(\angle YDB)=10 ^\circ$ . Let $ AC\cap BD=O$ . Then in $ \Delta AOD: m(\angle DAO)=10 ^\circ , m(\angle ADO)=80^\circ =>$ $ AC\bot BD$ .

Proof 3 (metric). $ ABCD$ quadrilater is orthodiagonal if and only if $ AB^2+CD^2=AD^2+BC^2\ (*)$ . I"ll use the Sinus' theorem :

$\left\{\begin{array}{cccc} \triangle ABC\ : & \frac {AB}{\sin 50^{\circ}}=\frac {BC}{\sin 30^{\circ}}=\frac {CA}{\sin 100 ^{\circ}} & \implies & \left\{\begin{array}{c} AB=\frac {AC}{2\cos 50^{\circ}}\\\\ BC=\frac {AC}{2\sin 80^{\circ}}\end{array}\right|\\\\ \triangle ADC\ : & \frac {AD}{\sin 20^{\circ}}=\frac {DC}{\sin 10^{\circ}}=\frac {CA}{\sin 30 ^{\circ}} & \implies & \left\{\begin{array}{c} AD=2\cdot AC\cdot\sin 20^{\circ}\\\\ DC=2\cdot AC\cdot \sin 10^{\circ}\end{array}\right|\end{array}\right|$ .

Replacing in $ (*)$ the last 4 relations we only have to prove that $ \boxed{\frac{1}{4\cos^2 50^{\circ}}+4\sin^2 10^{\circ}=4\sin^2 20^{\circ}+\frac{1}{4\sin^2 80^{\circ}}}$ .

$\boxed{\begin{array}{c} \\ \frac{1}{4\cos^2 50^{\circ}}+4\sin^2 10^{\circ}=4\sin^2 20^{\circ}+\frac{1}{4\sin^2 80^{\circ}}\\\\ \frac 14\cdot\left(\frac {1}{\sin^240^{\circ}}-\frac {1}{\sin^280^{\circ}}\right)=4\cdot\left(\sin^220^{\circ}-\sin^210^{\circ}\right)\\\\ \sin^280^{\circ}-\sin^240^{\circ}=16\sin^240^{\circ}\sin^280^{\circ}\left(\sin^220^{\circ}-\sin^210^{\circ}\right)\\\\ \sin40^{\circ}\sin120^{\circ}=16\sin^240^{\circ}\cos^210^{\circ}\sin 30^{\circ}\sin 10^{\circ}\\\\ \sin 60^{\circ}=4\sin 40^{\circ}\cos 10^{\circ}\sin 20^{\circ}\\\\ \sin 20^{\circ}\left(3-4\sin^220^{\circ}\right)=4\sin 40^{\circ}\cos 10^{\circ}\sin 20^{\circ}\\\\ 3-2(1-\cos 40^{\circ})=2(\sin 50^{\circ}+\sin 30^{\circ})\\\\ 1+2\cos 40^{\circ}=2\sin 50^{\circ}+1\\\\ \mathrm{O.K.}\\ \end{array}}$ .

I used the well-known identities $\left\{\begin{array}{c} x+y=180^{\circ}\implies \left\{\begin{array}{c} \sin x=\sin y\\\ \cos x=-\cos y\end{array}\right|\\\\ x+y=90^{\circ}\implies \sin x=\cos y\\\\ 2\sin x\cos x=\sin 2x\\\\ \sin 3x=\sin x(3-4\sin^2x)\\\\ \sin^2x-\sin^2y=\sin (x+y)\sin (x-y)\end{array}\right|$ .

Proof 4 (trigonometric - brute force). Denote $m\left(\widehat{ABD}\right)=x$ . Thus, $\left\{\begin{array}{c} m\left(\widehat{CBD}\right)=100^{\circ} -x\\\\ m\left(\widehat{ADB}\right)=140^{\circ}-x\\\\ m\left(\widehat{BDC}\right)=10^{\circ}+x\end{array}\right\|$ . Apply an well-known identity

$\boxed{\ \sin\widehat{ABD}\sin\widehat{DAC}\sin\widehat{CDB}\sin\widehat{BCA} =\sin\widehat{BAC}\sin\widehat{ADB}\sin\widehat{DCA}\sin\widehat{CBD}\ }$ . Therefpre,

$\sin x\sin 10^{\circ}\sin (10^{\circ}+x)\sin 50^{\circ}=\sin 30^{\circ}\sin (40^{\circ}+x)\sin 20^{\circ}\sin (80^{\circ}+x)\iff$

$2\sin x\sin 10^{\circ}\sin (10^{\circ}+x)\cos 40^{\circ}=\sin (40^{\circ}+x)\sin 20^{\circ}\cos (10^{\circ}-x)\iff$

$\sin x\sin (10^{\circ}+x)\cos 40^{\circ}=\sin (40^{\circ}+x)\cos 10^{\circ}\cos (10^{\circ}-x)\iff$

$\cos 40^{\circ}[\cos 10^{\circ}-\cos (10^{\circ}+2x)]=\cos 10^{\circ}[\sin 50^{\circ}+\sin (30^{\circ}+2x)]\iff$

$\cos 40^{\circ}\cos (10^{\circ}+2x)+\cos 10^{\circ}\sin (30^{\circ}+2x)=0\ \ (**)\ \iff$

$\cos 40^{\circ}(\cos 10^{\circ}-\sin 10^{\circ}\tan 2x)+\cos 10^{\circ}(\sin 30^{\circ}+\cos 30^{\circ}\tan 2x)=0\iff$

$\tan 2x=\frac {\cos 40^{\circ}\cos 10^{\circ}+\cos 10^{\circ}\sin 30^{\circ}}{-\cos 10^{\circ}\cos 30^{\circ}+\cos 40^{\circ}\sin 10^{\circ}}\iff$ $\tan 2x=\frac {\cos 50^{\circ}+\cos 30^{\circ}+\cos 10^{\circ}}{-\cos 40^{\circ}-\cos 20^{\circ}+\cos 40^{\circ}-\sin 30^{\circ}}\iff$

$\tan 2x=\frac {\cos 30^{\circ}+2\cos 30^{\circ}\cos 20^{\circ}}{-\cos 20^{\circ}-\frac 12}\iff$ $\tan 2x=\frac {2\cos 30^{\circ}(1+2\cos 20^{\circ})}{-(1+2\cos 20^{\circ})}\iff$ $\tan 2x=-\sqrt 3\iff x=60^{\circ}\iff AC\perp BD\iff$ $\mathrm{O.K.}$

Remark. From the step $(**)$ can continue thus :

$\cos 40^{\circ}\cos (10^{\circ}+2x)+\cos 10^{\circ}\sin (30^{\circ}+2x)=0\ \ (**)\ \iff$

$\cos (50^{\circ}+2x)+\cos (30^{\circ}-2x)+\cos (50^{\circ}-2x)+\cos (70^{\circ}-2x)=0\iff$

$ [\cos (50^{\circ}+2x)+\cos (50^{\circ}-2x)]+[\cos (30^{\circ}-2x)+\cos (70^{\circ}-2x)]=0\iff$

$\sin 40^{\circ}\cos 2x+\cos 20^{\circ}\cos (50^{\circ}-2x)=0\iff$ $2\sin 20^{\circ}\cos 2x+\cos (50^{\circ}-2x)=0\iff$

$\sin (20^{\circ}+2x)+\sin (20^{\circ}-2x)+\sin (40^{\circ}+2x)=0\iff$ $\cos (70^{\circ}-2x)+2\sin 30^{\circ}\cos (10^{\circ}+2x)=0\iff$

$ \cos (70^{\circ}-2x)=\cos (170-^{\circ}2x)\iff$ $(70^{\circ}-2x)+(170^{\circ}-2x)=0\iff$ $x=60^{\circ}\iff$ $AC\perp BD$

See and here ==> http://www.artofproblemsolving.com/blog/32104

GENERALIZATION. Let $ ABCD$ be a convex quadrilateral with $m(\angle DAC)=x$ , $m(\angle DCA)=30^{\circ}-x$ ,

$m(\angle BAC)=30^{\circ}$ and $ \angle BCA=60^{\circ}-x$ , where $0<x<30^{\circ}$ . Prove that $ m(\angle CBD)=4x$ .


PP5. Prove that $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{2\pi}{7}=\frac12$ .

Proof. Denote $z=\cos\phi +i\cdot\sin\phi $ . Then for any $n\in\mathbb N^*$ we have $\left\{\begin{array}{c} z^n=\cos n\phi +i\cdot\sin n\phi\\\\ \overline z^n=\frac {1}{z^n}=\cos n\phi -i\cdot\sin n\phi\end{array}\right\|\implies$ $\left\{\begin{array}{c} \cos n\phi =\frac {z^{2n}+1}{2z^n}\\\\ \sin n\phi =\frac {z^{2n}-1}{2iz^n}\end{array}\right\|$ .

In the particular case $\phi :=\frac {2\pi}{7}$ obtain that $z=\cos\frac {2\pi}{7} +i\cdot\sin\frac {2\pi}{7}$ and $\cos\frac {2n\pi}{7}=\frac {z^{2n}+1}{2z^n}$ . Observe that $z^7=1$ , i.e.

$\boxed{1+z+z^2+z^3+z^4+z^5+z^6=0}\ (*)$ . In conclusion, $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{2\pi}{7}=$ $-\cos\frac{2\pi}{7}-\cos\frac{4\pi}{7}-\cos\frac{6\pi}{7}=$

$-\left(\frac {z^2+1}{2z}+\frac {z^4+1}{2z^2}+\frac {z^6+1}{2z^3}\right)=$ $-\frac {z^2\left(z^2+1\right)+z\left(z^4+1\right)+\left(z^6+1\right)}{2z^3}=$ $-\frac {1+z+z^2+z^4+z^5+z^6}{2z^3}\stackrel{(*)}{=}$ $-\frac {-z^3}{2z^3}=\frac 12$ .

Remark. $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}-\cos\frac{2\pi}{7}=$ $2\cos\frac{2\pi}{7}\cos\frac{\pi}{7}-\cos \frac{2\pi}{7}=$ $-\cos \frac{2\pi}{7}\left(2\cos \frac{6\pi}{7}+1\right)=$

$-\frac {z^2+1}{2z}\left(2\cdot\frac {z^6+1}{2z^3}+1\right)=$ $-\frac {\left(z^2+1\right)\left(z^6+z^3+1\right)}{2z^4}=$ $-\frac {z+z^5+z^2+z^6+z^3+1}{2z^4}=$ $-\frac {-z^4}{2z^4}=\frac 12$ .


PP6. Show that $\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}$ , where $n\in\mathbb N$ , $n\ge 2$ .

Proof. If denote $w=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$ , then $\left\{w,w^2,\ \cdots\ ,\omega^{n-1}\right\}$ are the roots of the equation $x^{n-1}+x^{n-2}+\cdots+x+1=0$ ,

i.e. $x^{n-1}+x^{n-2}+\cdots+x+1=\prod_{k=1}^{n-1}\left(x-w^k\right)$ . Plugging $x:=1$ obtain that $n=\prod_{k=1}^{n-1}\left(1-w^k\right)$ $\Longrightarrow$ $ n=\left|\prod_{k=1}^{n-1}\left(1-w^k\right)\right|$ , where

$\left|1-w^k\right|=\left|1-\cos\frac{2k\pi}{n}-i\sin\frac{2k\pi}{n}\right|=2\sin\frac{k\pi}{n}$ , where $k\in\overline{0,n-1}$ . Thus, $2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=n$ $\Longrightarrow\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac{n}{2^{n-1}}$ .

Remark. A geometrical interpretation of this identity. Let $A_1A_2\ \ldots\ A_n$ be a regular polygon inscribed in a circle with radius $1$ . Then $A_1A_2\cdot A_1A_3\cdot\ldots\cdot A_1A_n=n$ .

The new identity $\boxed{\ \prod_{k=1}^{n}\sin\frac {(2k-1)\pi}{2n}=\frac{1}{2^{n-1}}\ }$ is a paricular case of the well-known identity $\boxed{\prod_{k=1}^{n-1}\sin \frac{k \pi}n = \frac n{2^{n-1}}}$ for $n:=2n$ .

Indeed, $\prod_{k=1}^{2n-1}\sin\frac {k\pi}{2n}=\frac {2n}{2^{2n-1}}\iff$ $\prod_{k=1}^{n}\sin\frac {(2k-1)\pi}{2n}\cdot\prod_{k=1}^{n-1}\sin\frac {k\pi}{n}=\frac {n}{2^{2(n-1)}}\iff$ $\prod_{k=1}^{n}\sin\frac {(2k-1)\pi}{2n}=\frac{1}{2^{n-1}}$ .

GENERALIZATION. $f_n(x)\equiv \prod_{k=1}^{n-1}\sin\left(x+\frac {k\pi}{n}\right)=\frac {1}{2^{n-1}}\cdot\frac {\sin nx}{\sin x}$ . Thus, $\lim_{x\to 0}f_n(x)=\frac {n}{2^{n-1}}\implies$ $g_n(x)=\left\{\begin{array}{ccc} f_n(x) & \mathrm{if} & x\ne 0\\\\ \frac {n}{2^{n-1}} & \mathrm{if} & x=0\end{array}\right\|$ is a continue function.


PP7. Prove that $\cos^7{x}+\cos^7\left(x+\frac{2\pi}{3}\right)+\cos^7\left(x+\frac{4\pi}{3}\right)=\frac{63}{64}\cdot\cos{3x}$ , where $x\in\mathbb R$ .

Proof 1 (with complex numbers). $w=\cos\frac {2\pi}{3}+i\cdot\sin\frac {2\pi}{3}\ \implies\ w^3=1$ , $w^2+w+1=0$ and $\overline w=\frac {1}{w}=w^2\ ;$

$z=\cos x+i\cdot\sin x\ \implies$ for any $n\in \mathbb N$ we have $\cos nx=\frac {z^{2n}+1}{2z^n}$ . Thus, $\left\{\begin{array}{ccc} \cos x=\frac {z^2+1}{2z} & ; & \cos 3x=\frac {z^6+1}{2z^3}\\\\ \cos\left(x+\frac {2\pi}{3}\right)=\frac {w^2z^2+1}{2wz} & ; & \cos\left(x+\frac {4\pi}{3}\right)=\frac {z^2+w^2}{2wz}\end{array}\right\|$ .

Therefore, our identity is equivalently with an identity in $\mathbb C[X]\ :\ \left(\frac {z^2+1}{2z}\right)^7+\left(\frac {w^2z^2+1}{2wz}\right)^7+\left(\frac {z^2+w^2}{2wz}\right)^7=\frac {63}{64}\cdot\frac {z^6+1}{2z^3}\iff$

$w\left(z^2+1\right)^7+\left(w^2z^2+1\right)^7+\left(z^2+w^2\right)^7=63wz^4\left(z^6+1\right)$ , what is truly because $w^{3k}=1$ for any $k\in\mathbb Z$ and $w^2+w+1=0$ .

Proof 2 (with fundamental symmetrical forms). Denote $\left\{\begin{array}{ccc} a & = & \cos x\\\\ b & = & \cos\left(x+\frac{2\pi}{3}\right)\\\\ c & = & \cos\left(x+\frac{4\pi}{3}\right)\end{array}\right\|$ . Prove easily that

$\left\{\begin{array}{ccc} s_1=a+(b+c)=\cos x+2\cos (x+\pi )\cos\frac {\pi}{3}=\cos x-\cos x & \implies & \boxed {s_1=0}\\\\ s_2=bc+a(b+c)=bc-a^2\implies 2s_2=\cos (2x+2\pi )+\cos\frac {2\pi}{3}-(1+\cos 2x) & \implies & \boxed{s_2=-\frac 34}\\\\ s_3=a(bc)\implies 4s_3=2\cos x\left[\cos (2x+2\pi )+\cos\frac {2\pi}{3}\right]=2\cos x\cos 2x-\cos x & \implies & \boxed{s_3=\frac 14\cdot\cos 3x}\end{array}\right\|$ .

Therefore,$\{a,b,c\}$ are the roots of the equation $\boxed{4t^3-3t-\cos 3x=0\begin{array}{c} \nearrow\\\ \rightarrow\\\ \searrow\end{array}\begin{array}{c} a\\\\ b\\\\ c\end{array}}$ . Denote $S_n=a^n+b^n+c^n$ , where $n\in\mathbb N$ .

Show easily that for any $n\in\mathbb N$ we have $\boxed{4\cdot S_{n+3}=3\cdot S_{n+1}+\cos 3x\cdot S_n}$ , where $S_0=3$ , $S_1=0$ and $S_2=\frac 32$ . Thus,

$\underline{64\cdot S_7}=16\cdot\left(3\cdot S_5+\cos 3x\cdot S_4\right)=$ $12\cdot\left(3\cdot S_3+\cos 3x\cdot S_2\right)+$ $4\cos 3x\cdot \left(3\cdot S_2+\cos 3x\cdot S_1\right)=$

$9\cdot\left(3\cdot S_1+\cos 3x\cdot S_0\right)+$ $24\cos 3x\cdot S_2+4\cos^23x\cdot S_1=$ $\underline{27\cos 3x+36\cos 3x}$ $\implies$ $\boxed{S_7=\frac {63}{64}\cdot\cos 3x}$ .


Lemma. Consider the real numbers $r\ne 0$ , $x_1$ and $x_{k+1}=x_k+r$ , where $k\in\mathbb N^*$ .

Then $\boxed{C\equiv\sum_{k=1}^n\cos x_k=\frac {\cos\frac {x_1+x_n}{2}\sin\frac {nr}{2}}{\sin\frac r2}}$ and $\boxed{S\equiv\sum_{k=1}^n\sin x_k=\frac {\sin\frac {x_1+x_n}{2}\sin\frac {nr}{2}}{\sin\frac r2}}$ .

Proof. $2\sin\frac r2\cdot C=\sum_{k=1}^n2\sin\frac r2\cos x_k=$ $\sum_{k=1}^n\left[\sin \left(x_k+\frac r2\right)-\sin \left(x_k-\frac r2\right)\right]=$

$\sin \left(x_n+\frac r2\right)-\sin \left(x_1-\frac r2\right)$ , because $x_{k+1}-\frac r2=x_k+\frac r2$ . In conclusion, $2\sin\frac r2\cdot C=$

$2\sin \frac {x_n-x_1+r}{2}\cos\frac {x_1+x_n}{2}=$ $2\sin \frac {(n-1)r+r}{2}\cos\frac {x_1+x_n}{2}\implies$ $\boxed{C=\frac {\cos\frac {x_1+x_n}{2}\sin\frac {nr}{2}}{\sin\frac r2}}$ .

$\boxed{2\sin\frac r2\cdot S}=\sum_{k=1}^n2\sin\frac r2\sin x_k=$ $\sum_{k=1}^n\left[\cos \left(x_k-\frac r2\right)-\cos\left(x_k+\frac r2\right)\right]=$

$\cos\left(x_1-\frac r2\right)-\cos\left(x_n+\frac r2\right)$ , because $x_{k+1}-\frac r2=x_k+\frac r2$ . In conclusion, $2\sin\frac r2\cdot S=$

$2\sin \frac {x_n-x_1+r}{2}\sin\frac {x_1+x_n}{2}=$ $2\sin \frac {(n-1)r+r}{2}\sin\frac {x_1+x_n}{2}\implies$ $\boxed{S=\frac {\sin\frac {x_1+x_n}{2}\sin\frac {nr}{2}}{\sin\frac r2}}$ .

PP8. Denote $E\equiv\sum_{k=1}^n\sin^2\frac{k \pi m}{n}$ , where $m,n \in \mathbb{Z}$ and $0<m<n$ . Find $E$ as a function of $n$ .

Proof. $2E=\sum_{k=1}^n\left(1-\cos\frac{2k \pi m}{n}\right)=n-F$ , where $F\equiv \sum_{k=1}^n\cos\frac{2k \pi m}{n}$ . Apply upper lemma for $x_1=r=\frac {2m\pi}{n}$ and

obtain that $F=\frac {\cos\frac {x_1+x_n}{2}\sin\frac {nr}{2}}{\sin\frac r2}=$ $\frac {\cos \frac {x_1+x_n}{2}\sin \left(\frac n2\cdot\frac {2m\pi}{n}\right)}{\sin \frac r2}=0\implies$ $F=0$ . In conclusion, $2E=n\implies$ $E=\frac n2$ .


PP9. Prove that $S_n\equiv\sum_{k=1}^{n}(-1)^{k+1}\cos \dfrac{k\pi}{2n + 1} = \dfrac{1}{2}$ and $P_n\equiv\prod_{k=1}^{n}\cos \dfrac{k\pi}{2n + 1} = \dfrac{1}{2^n}$ .

Proof. I"ll use the remarkable identity $(\forall )\ k\in\overline{1,n}\ ,\ a_{k+1}=a_k+r\ \implies \boxed{\sum_{k=1}^n\cos a_k=\frac {\cos\frac {a_1+a_n}{2}\sin\frac {nr}{2}}{\sin \frac r2}}$ . Indeed,

$S_n=\sum_{k=1}^n\cos\left[\frac {k\pi}{2n+1}+(k-1)\pi\right]=$ $\frac {\cos\frac {n^2\pi}{2n+1}\sin\frac {n(n+1)\pi}{2n+1}}{\sin\frac {(n+1)\pi}{2n+1}}=$ $\frac {\sin\frac {n(2n+1)\pi}{2n+1}+\sin\frac {n\pi}{2n+1}}{2\sin\frac {(n+1)\pi}{2n+1}}=$ $\frac {\sin\frac {n\pi}{2n+1}}{2\sin\frac {(n+1)\pi}{2n+1}}=\frac 12$

because $\sin n\pi =0$ and $\frac {n\pi}{2n+1}+\frac {(n+1)\pi}{2n+1}=\pi$ . Remark that $\boxed{\sum_{k=1}^n\sin a_k=\frac {\sin\frac {a_1+a_n}{2}\sin\frac {nr}2}{\sin \frac r2}}$ .

Denote $w=\cos\frac {2\pi}{2n+1}+i\cdot\sin\frac {2\pi}{2n+1}$ . Observe that the roots of the equation $z^{2n+1}=1$ are $w^k$ , whee $k\in\overline{0,2n}$ , i.e. $\sum_{k=0}^{2n}z^k=\prod_{k=1}^{2n}\left(z-w^k\right)$ .

For $z:=-1$ obtain that $1=\prod_{k=1}^{2n}\left(1+w^k\right)=$ $\prod_{k=1}^{2n}\left(1+\cos\frac {2k\pi}{2n+1}+i\cdot\sin\frac {2k\pi}{2n+1}\right)=$ $\prod_{k=1}^{2n}\left(2\cos^2\frac {k\pi}{2n+1}+2i\cdot \cos\frac {k\pi}{2n+1}\sin\frac {k\pi}{2n+1}\right)=$

$4^n\cdot\prod_{k=1}^{2n}\cos\frac {k\pi}{2n+1}\cdot\prod_{k=1}^{2n}\left(\cos\frac {k\pi}{2n+1}+i\cdot \sin\frac {k\pi}{2n+1}\right)=$ $4^n\cdot\prod_{k=1}^{n}\cos\frac {k\pi}{2n+1}\cdot$ $\prod_{k=n+1}^{2n}\cos\frac {k\pi}{2n+1}\cdot$

$\left(\cos\frac {\pi}{2n+1}\cdot\sum_{k=1}^{2n}k+i\cdot \sin\frac {\pi}{2n+1}\cdot\sum_{k=1}^{2n}k\right)=$ $4^n\cdot\prod_{k=1}^n\cos\frac {k\pi}{2n+1}\cdot (-1)^n\prod_{k=1}^n\cos\frac {k\pi}{2n+1}\cdot (-1)^n\implies$ $\boxed{\prod_{k=1}^n\cos\frac {k\pi}{2n+1}=\frac {1}{2^n}}$ .

Remark. I used the evident relations $\prod_{k=m}^nf(k)=$ $\prod_{k=1}^{n-m+1}f(k+m-1)=$ $\prod_{k=1}^{n-m+1}f(n+1-k)$ of the product/sum.


PP10. Ascertain $\cos 36^{\circ}-\cos 72^{\circ}$ .

Proof 0 (trigonometric - pco's). Denote $S=1+\cos \frac{2\pi}5$ $+\cos \frac{4\pi}5$ $+\cos \frac{6\pi}5+\cos \frac{8\pi}5$ . Observe that

$2\sin\frac {\pi}{5}\cdot S=$ $2\sin\frac {\pi}{5}+\left(\sin\frac {3\pi}{5}-\sin\frac {\pi}{5}\right)+$ $\left(\sin\frac {5\pi}{5}-\sin\frac {3\pi}{5}\right)+$ $\left(\sin\frac {7\pi}{5}-\sin\frac {5\pi}{5}\right)+$ $\left(\sin\frac {9\pi}{5}-\sin\frac {7\pi}{5}\right)=$

$\sin \frac {\pi}{5}+\sin\frac {9\pi}{5}=\sin \frac {\pi}{5}-\sin\frac {\pi}{5}=0$ . Thus, $S=0\implies 1+2\cos \frac{2\pi}5-2\cos \frac{\pi}5=0$ $\implies$ $\boxed{\cos \frac{\pi}5-\cos \frac{2\pi}5=\frac 12}$ .

Remark. Let the arthmetrical progression $\alpha_k\in\mathbb R\ ,\ k\in\overline{1,n}$ , i.e. $\alpha_{k+1}=\alpha_k+r$ , where $r\in \mathbb R$ (constant).

Prove easily that $\sum_{k=1}^n\sin\alpha_k=\frac {\sin\frac {\alpha_1+\alpha_n}2\sin\frac {nr}2} {\sin\frac r2}$ and $\sum_{k=1}^n\cos\alpha_k=\frac {\cos\frac {\alpha_1+\alpha_n}2\sin\frac {nr}2}{\sin\frac r2}$ .

Proof 1 (trigonometric). Observe that $2\sin 36^{\circ}\cdot \left(\cos 36^{\circ}-\cos 72^{\circ}\right)=$ $2\sin 36^{\circ}\cos 36^{\circ}-2\sin 36^{\circ}\cos 72^{\circ}=$

$\sin 72^{\circ}-\left(\sin 108^{\circ}-\sin 36^{\circ}\right)\implies$ $2\sin 36^{\circ}\cdot \left(\cos 36^{\circ}-\cos 72^{\circ}\right)=\sin 36^{\circ}\implies$ $\boxed{\cos 36^{\circ}-\cos 72^{\circ}=\frac 12}$ .

Proof 2 (geometric). Consider an $A$-isosceles $\triangle ABC$ with $A=36^{\circ}$ . Denote the point $D\in (AC)$ for which $\widehat{DBA}\equiv\widehat{DBC}$ .

Observe that $AD=DB=BC$ . Let $E\ ,\ F$ be the midpoints of the segments $[AB]\ ,[CD]$ respectively. Thus, $ABC\sim BCD$

$\implies$ $\frac {AB}{BC}=\frac{BC}{CD}\ (*)$ . Observe that $\cos 36^{\circ}=\frac {AE}{AD}=\frac {AB}{2\cdot AD}=\frac {AC}{2\cdot AD}$ and $\cos 72^{\circ}=\frac {DF}{DB}=\frac {DF}{AD}=\frac {DC}{2\cdot AD}$ .

In conclusion, $\cos 36^{\circ}-\cos 72^{\circ}=$ $\frac {AC}{2\cdot AD}-\frac {DC}{2\cdot AD}=$ $\frac {AC-DC}{2\cdot AD}=\frac {AD}{2\cdot AD}\implies$ $\boxed{\cos 36^{\circ}-\cos 72^{\circ}=\frac 12}$ .

Remark. Suppose w.l.o.g. $AB=1$ and denote $AD=DB=BC=x$ . Thus, $DC=1-x$ . From the relation $(*)$ obtain that

$x^2=1-x\iff$ $ x^2+x-1=0$ , i.e. $x=\frac {-1+\sqrt 5}{2}$ and $\cos 36^{\circ}=\frac {AE}{AD}=\frac {1}{2x}=\frac {1}{-1+\sqrt 5}\implies$ $\boxed{\cos 36^{\circ}=\frac {1+\sqrt 5}{4}}$ .
$\cos 72^{\circ}=2\cos^236^{\circ}-1=$ $2\cdot \frac {6+2\sqrt 5}{16}-1=$ $\frac {3+\sqrt 5}{4}-1\implies$ $\boxed{\cos 72^{\circ}= \frac {-1+\sqrt 5}{4}}$ . So $\boxed{\cos 36^{\circ}-\cos 72^{\circ}=\frac 12}$ .


PP11. Prove that $E\equiv \tan x\tan(x+60^{\circ})+\tan x\tan(x-60^{\circ})+\tan(x+60^{\circ}) \tan(x-60^{\circ})$ is constant for any $x$ so that $\left\{x,x\pm \frac {\pi}{3}\right\}\cap \left(\pi Z+\frac {\pi}{2}\right)=\emptyset$ .

Proof 1. I"ll use the well-known conditioned identity $\boxed{(\alpha +\beta +\gamma)\in\pi Z\implies \tan \alpha +\tan\beta +\tan\gamma=\tan \alpha\cdot\tan\beta\cdot\tan\gamma}\ :$

$\left\{\begin{array}{ccc} \left|\begin{array}{ccc} \alpha & = & 60^{\circ}\\\\ \beta & = & x+60^{\circ}\\\\ \gamma & = & -x+60^{\circ}\end{array}\right| & \implies & \sqrt 3+\tan (x+60^{\circ})-\tan (x-60^{\circ})=-\sqrt 3\tan (x+60^{\circ})\tan (x-60^{\circ})\\\\ \left|\begin{array}{ccc} \alpha & = & 60^{\circ}\\\\ \beta & = & -x-60^{\circ}\\\\ \gamma & = & x\end{array}\right| & \implies & \sqrt 3-\tan (x+60^{\circ})+\tan x=-\sqrt 3\tan (x+60^{\circ})\tan x\\\\ \left|\begin{array}{ccc} \alpha & = & 60^{\circ}\\\\ \beta & = & x-60^{\circ}\\\\ \gamma & = & -x\end{array}\right| & \implies & \sqrt 3+\tan (x-60^{\circ})-\tan x=-\sqrt 3\tan (x-60^{\circ})\tan x\end{array}\right\|\bigoplus$ $\implies\ 3\sqrt 3=-E\sqrt 3\implies E=-3$ .

Proof 2. I"ll use the remarkable identity $\boxed{\sin^2a-\sin^2b=\sin (a+b)\sin (a-b)}$ . Thue, $E=\tan x\left[\tan (x+60)+\tan (x-60)\right]+\tan (x+60)\tan (x-60)=$

$\frac {\sin x}{\cos x}\cdot\frac {\sin [(x+60)+(x-60)]}{\cos (x+60)\cos (x-60)}+\frac {\sin (x+60)\sin (x-60)}{\cos (x+60)\cos (x-60)}=$ $\frac {2\sin^2x+\sin^2x-\sin^260}{\sin (30+x)\sin (30-x)}=$ $\frac {2\sin^2x+\sin^2x-\sin^260}{\sin^230-\sin^2x}=$ $\frac {3\left(4\sin^2x-1\right)}{1-4\sin^2x}=-3$ .

Proof 3 (common). Denote $\tan x=t$ . Thus, $E=\tan x\left[\tan (x+60)+\tan (x-60)\right]+\tan (x+60)\tan (x-60)=$ $t\cdot\left(\frac {t+\sqrt 3}{1-t\sqrt 3}+\frac {t-\sqrt 3}{1+t\sqrt 3}\right)+$

$\frac {t+\sqrt 3}{1-t\sqrt 3}\cdot\frac {t-\sqrt 3}{1+t\sqrt 3}=$ $\frac {t\left[(t+\sqrt 3)(1+t\sqrt 3)+(1-t\sqrt 3)(t-\sqrt 3)\right]+\left(t^2-3\right)}{1-3t^2}=$ $\frac {t\cdot 8t+\left(t^2-3\right)}{1-3t^2}=$ $\frac {9t^2-3}{1-3t^2}=$ $\frac {3\left(3t^2-1\right)}{1-3t^2}=-3$ .


PP12. Show that: $ \frac{1}{sin 10^{\circ} }-\frac{\sqrt 3}{cos10^{\circ}}=4$ .

Proof. $ \frac{1}{sin 10^{\circ} }-\frac{\sqrt 3}{cos10^{\circ}}=4\iff$ $\cos 10^{\circ}-\sin 10^{\circ}\sqrt 3=2\sin 20^{\circ}\iff$ $\cos 60^{\circ}\cos 10^{\circ}-\sin 60^{\circ}\sin 10^{\circ}=\sin 20^{\circ}\iff$ $\cos 70^{\circ}=\sin 20^{\circ}$ , what is truly.


PP13. Ascertain $S_n(x)=\sum_{k=1}^n kx^k$ , where $x\in\mathbb C$ and for $x\in (0,1)$ , find the limit $S(x)=\lim_{n\to\infty}\sum_{k=1}^n kx^k$ .

Proof 1. I"ll use the well-known property of the symbol $\sum\ :\ \sum_{k=m}^nT_k=\sum _{k=m+p}^{n+p}T_{k-p}$ , where $\{m,n,p\}\subset\mathbb Z$ . $S_n(x)=\sum_{k=1}^nkx^k=\sum_{k=0}^{n-1}(k+1)x^{k+1}=$ $x\cdot \sum _{k=0}^{n-1}kx^k+\sum_{k=0}^{n-1}x^{k+1}=$

$x\cdot\left(\sum_{k=1}^nkx^k-nx^n\right)+\sum_{k=1}^nx^k=$ $x\cdot \left[S_n(x)-nx^n\right]+\frac {x\cdot\left(1-x^n\right)}{1-x}\implies$ $S_n(x)=\frac {-nx^{n+1}+\frac {x\cdot\left(1-x^n\right)}{1-x}}{1-x}\implies$ $\boxed{S_n(x)=\frac {nx^{n+2}-(n+1)x^{n+1}+x}{(1-x)^2}}$ .

In conclusion, $x\in (0,1)\implies \lim_{n\to\infty} nx^n=0$ and if denote $S(x)\equiv \lim_{n\to\infty}S_n(x)$ , then obtain that $\boxed{S(x)=\frac {x}{(1-x)^2}}=\frac {1}{(1-x)^2}-\frac {1}{1-x}$ .

Proof 2. $S_n(x)=\sum_{k=1}^nkx^k=$ $x\cdot\left(\sum_{k=1}^nx^k\right)'=$ $x\cdot\left(\frac {x^{n+1}-x}{x-1}\right)'\implies$ $S_n(x)=x\cdot \frac {\left[(n+1)x^n-1\right](x-1)-\left(x^{n+1}-x\right)}{(x-1)^2}$ $\implies$ $\boxed{S_n(x)=\frac {x\cdot\left[nx^{n+1}-(n+1)x^n+1\right]}{(x-1)^2}}$ .

Proof (3). $S_n(x)=\sum_{k=1}^nkx^k=$ $\sum_{s=1}^n\sum_{k=s}^nx^k=$ $\sum_{k=1}^n\frac {x^s\left(x^{n+1-s}-1\right)}{x-1}=$ $\frac {1}{x-1}\cdot \left(nx^{n+1}-\sum_{s=1}^nx^s\right)=$ $\frac {1}{x-1}\cdot \left(nx^{n+1}-\frac {x^{n+1}-x}{x-1}\right)\implies$ $\boxed{S_n(x)=\frac {nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}}$ .

Remark. Obtain (or with the Horner's schema) that $\boxed{nx^{n+1}-(n+1)x^n+1=(x-1)^2\cdot\left[\ 1+2x+3x^2+\ \ldots\ +(n-1)x^{n-2}+nx^{n-1}\right]\ }\ (*)$

and $nx^{n+1}-(n+1)x^n+1=nx^n(x-1)-\left(x^n-1\right)=$ $(x-1)\left(nx^n-x^{n-1}-x^{n-2}-\ \ldots\ -x^2-x-1\right)\implies$

$\boxed{\ nx^n-x^{n-1}-x^{n-2}-\ \ldots\ -x^2-x-1=(x-1)\left[nx^{n-1}+(n-1)x^{n-2}+\ \ldots\ +3x^2+2x+1\right]\ }$ .

From the relation $(*)$ obtain that $x>0\implies \frac {nx^{n+1}+1}{n+1}\ge x^n$ , with the equality iff $x=1$ . Using the substitution $x:=\sqrt [n+1]a\ ,\ a>0$ obtain that the inequality $\frac {1+na}{n+1}\ge \sqrt[n+1]{a^n}\ (1)$ , where $a>0$ ,

i.e. the A.M./G.M.-inequality in the case $a_k=a$ for any $k\in \overline {1,n}$ and $a_{n+1}=1$ . Let $a_k\ ,\ k\in\mathbb N^*$ be positive numbers for which denote $n\cdot A_n=\sum_{k=1}^n a_k$ , where $A_n$ is the arithmetical mean and $G_n^n=\prod_{k=1}^na_k$ ,

where $G_n$ is the geometrical mean. Using the inequality $(1)$ for $a:=\frac {G_n}{a_{n+1}}$ obtain that $1+n\cdot\frac {G_n}{a_{n+1}}\ge (n+1)\cdot\sqrt [n+1]{\frac {G_n^n}{a_{n+1}^n}\cdot\frac {a_{n+1}}{a_{n+1}}}\iff$ $a_{n+1}+n\cdot G_n\ge (n+1)\cdot G_{n+1}\iff$

$(n+1)\cdot A_{n+1}-n\cdot A_n\ge (n+1)\cdot G_{n+1}-n\cdot G_n\iff$ $\boxed{\ (n+1)\cdot\left(A_{n+1}-G_{n+1}\right)\ \ge\ n\cdot\left(A_n-G_n\right)\ }$ (T. Popoviciu's inequality).