What is ∑ n²/2^n?

Question

In other words, the sum of the infinite series 1/2 + 4/4 + 9/8 + 16/16 + 25/32 + 36/64 + ....

Answer 1

The answer is 6. Check out this integer series, look under "formula"

Answer 2

The sum of ∑ n²/2^n from n = 1 to infinity is 6.

Since

x² d²/dx² [1/(1-x)] + x d/dx [1/(1-x)]

=2x²/(1-x)³ + x/(1-x)²

Substituting x = 1/2 give the expression equal to 6.

Answer 3

f(x) = ∑x^n = 1/(1-x)
f'(x) = ∑nx^(n-1) = d/dx [1/(1-x)]
f''(x) =∑n(n-1)x^(n-2) = d²/dx² [1/(1-x)]

f''(x) = ∑(n²-n)x^(n-2) = d²/dx² [1/(1-x)]

x²f''(x) = ∑(n²-n)x^n = x² d²/dx² [1/(1-x)]
xf'(x) = ∑nx^n = x d/dx [1/(1-x)]
x²f''(x) + xf'(x) = ∑n²x^n = x² d²/dx² [1/(1-x)] + x d/dx [1/(1-x)]


Answer:
∑n²(1/2)^n = ∑n²x^n =
= x² d²/dx² [1/(1-x)] + x d/dx [1/(1-x)] =
= (x³ + x)/(1-x)³ =
= 5