Answer:
∑₂°° n (n − 1) xⁿ⁻²
Explanation:
f(x) = 1 / (1 − x)
f'(x) = 1 / (1 − x)²
f''(x) = 2 / (1 − x)³
Therefore:
g(x) = f''(x)
g(x) = d²/dx² [1 / (1 − x)]
Using sum of a geometric series:
g(x) = d²/dx² ∑₀°° xⁿ
g(x) = d/dx ∑₁°° n xⁿ⁻¹
g(x) = ∑₂°° n (n − 1) xⁿ⁻²