Question

How do you find the value of `\sum _ { n } ^ { \infty } \frac { n ^ { 2} } { 5^ { n } } ( x - 1) ^ { n }`?

Answer 1

See below.

Explanation:

`sum_(k=0)^oo k^2y^k = yd/(dy)(yd/(dy)(sum_(k=0)^oo y^k))`

now making `y = (x-1)/5` and assuming `abs y < 1` we have

`sum_(k=0)^oo y^k = 1/(1-y)` and then

`sum_(k=0)^oo k^2y^k =-(y (1 + y))/(y-1)^3` or

`\sum _ { n } ^ { \infty } \frac { n ^ { 2} } { 5^ { n } } ( x - 1) ^ { n } = (5 (1 - x) (4 + x))/(x-6)^3`