利用:C(n,1)+2*C(n,2).....証明

(1) C(n,1)+2*C(n,2)+3*C(n,3)+...+n*C(n,n)=n*2^(n-1)

By A.M. >= G.M.

[C(n,1)+2*C(n,2)+3*C(n,3)+...+n*C(n,n)]/n >= [C(n,1)*2*C(n,2)*3*C(n,3)+...+n*C(n,n)]^(1/n)

So, [C(n,1)+2*C(n,2)+3*C(n,3)+...+n*C(n,n)]/n >= [n! C(n,1)*C(n,2)*C(n,3)...C(n,n)]^(1/n)

2^(n-1) >= [n! C(n,1)*C(n,2)*C(n,3)...C(n,n)]^(1/n)

C(n,1)*C(n,2)*C(n,3)...C(n,n) <= 2^[n(n-1)]/n!

Since C(n,1), 2*C(n,2), ... n*C(n,n) are not equal, the equality sign can be released and get the more strict result.

2 C(n,1)+2*C(n,2)+3*C(n,3)+...+n*C(n,n)=n*2^(n-1)

(1 + x)^n = Σ (nCk)x^k

differentiate two times on both sides w.r.t x

n(n - 1)(1 + x)^(n - 2) = Σ k(k-1)(nCk)x^(k - 2)

Sub. x = 1, n(n - 1)2^(n - 2) = Σ k(k-1)(nCk)

Now, Σ k^2(nCk)

= Σ k(k-1)(nCk) + Σ k(nCk)

= n(n - 1)2^(n - 2) + n*2^(n - 1)

= n*2^(n - 2)(n - 1 + 2)

= n(n + 1)2^(n - 2)