求级数∑（n=1,∞）n^2 / (2^(n-1))

f(x) = 1/(1-x) = 1+x+x^2+x^3+ .+x^n + ...
f'(x) = 1/(1-x)^2 = 1 + 2x + 3x^2 + ... + nx^(n-1) + (n+1)x^n + .
f''(x) = 2/(1-x)^3 = 2 + 6x + . + n(n-1)x^(n-2) + (n+1)nx^(n-1) + .
所以f'(x) + xf''(x) = 1 + 4x + 9x^2 + . + n^2 *x^(n-1) + .
令x=1/2得
原式 = 12