Let G(x) = 1/(1 - x)^2 = Σ_{i=0}^∞ g(i)x^i, where |x| < 1. What is g(i)?
2005
Let G(x) = 1/(1 - x)^2 = Σ_{i=0}^∞ g(i)x^i, where |x| < 1. What is g(i)?
- A.
i
- B.
i+1
- C.
2i
- D.
2^i
Show answer & explanation
Correct answer: B
Key idea: use the geometric series and differentiate.
Start from the geometric series valid for |x| < 1:
1/(1 - x) = ∑_{n=0}^∞ x^n.
Differentiate both sides with respect to x:
d/dx [1/(1 - x)] = d/dx [∑_{n=0}^∞ x^n].
The left-hand side becomes 1/(1 - x)^2. The right-hand side becomes ∑_{n=1}^∞ n x^{n-1}, which after reindexing (set i = n - 1) gives ∑_{i=0}^∞ (i + 1) x^i.
Therefore 1/(1 - x)^2 = ∑_{i=0}^∞ (i + 1) x^i, so the coefficient g(i) = i + 1.
Example: g(0) = 1, g(1) = 2, g(2) = 3.