n元恒等式

hbghlyj

$k=2$:
$$(u+v)^1=u^1+v^1$$
$k=3$:
$$(u+v+w)^2=(u+v)^2 + (v+w)^2 +(w+u)^2 -u^2 -v^2-w^2$$
$k=4$:
$$(u+v+w+x)^3 = (u+v+w)^3 + (u+v+x)^3 + (u+x+w)^3 + (x+v+w)^3 - (u+v)^3 - (v+w)^3 - (w+x)^3 - (u+w)^3 - (x+u)^3 -(v+x)^3 + u^3 +v^3+w^3+x^3$$
求证n元恒等式

hbghlyj

相关：@kuing的帖

hbghlyj

$n=1: a = a $   
$n=2: (a+b)^2 - a^2 - b^2 = 2ab$   
$n=3: (a+b+c)^3 - (a+b)^3-(b+c)^3-(c+a)^2 +a^3+b^3+c^3 = 6abc$
$n=4: (a+b+c+d)^4 - (a+b+c)^4-(b+c+d)^4-(c+d+a)^4-(d+a+b)^4 + (a+b)^4+(b+c)^4+(c+d)^4+(d+a)^4+(a+c)^4+(b+d)^4 - a^4-b^4-c^4-d^4 = 24abcd$