Question

`sum_(k=0)^ook^2 6^(-k) =` ?

Answer 1

`sum_(n=1)^oo n^2/6^n=42/125`

Explanation:

Taking this step by step, let us start with:

`5/6 sum_(n=1)^oo 1/6^n=sum_(n=1)^oo 1/6^n - 1/6 sum_(n=1)^oo 1/6^n`

`=1/6 + sum_(n=2)^oo 1/6^n - sum_(n=1)^oo 1/6^(n+1)`

`=1/6 + color(red)(cancel(color(black)(sum_(n=1)^oo 1/6^(n+1)))) - color(red)(cancel(color(black)(sum_(n=1)^oo 1/6^(n+1))))`

`=1/6`

Multiplying both ends by `6/5` we find:

`sum_(n=1)^oo 1/6^n = 1/5`

Building on this, we find:

`5/6 sum_(n=1)^oo n/6^n = sum_(n=1)^oo n/6^n - 1/6 sum_(n=1)^oo n/6^n`

`=sum_(n=1)^oo n/6^n - sum_(n=1)^oo n/6^(n+1)`

`=1/6 + sum_(n=2)^oo n/6^n - sum_(n=1)^oo n/6^(n+1)`

`=1/6 + sum_(n=1)^oo (n+1)/6^(n+1) - sum_(n=1)^oo n/6^(n+1)`

`=1/6 + sum_(n=1)^oo ((n+1) - n)/6^(n+1)`

`=1/6 + sum_(n=1)^oo 1/6^(n+1)`

`=1/6 + 1/6 sum_(n=1)^oo 1/6^n`

`=1/6 + 1/6(1/5)`

`=1/5`

Multiplying both ends by `6/5` we find:

`sum_(n=1)^oo n/6^n = 6/25`

Similarly, we find:

`5/6 sum_(n=1)^oo n^2/6^n=sum_(n=1)^oo n^2/6^n - 1/6 sum_(n=1)^oo n^2/6^n`

`=sum_(n=1)^oo n^2/6^n - sum_(n=1)^oo n^2/6^(n+1)`

`=1/6+sum_(n=2)^oo n^2/6^n - sum_(n=1)^oo n^2/6^(n+1)`

`=1/6+sum_(n=1)^oo (n+1)^2/6^(n+1) - sum_(n=1)^oo n^2/6^(n+1)`

`=1/6+sum_(n=1)^oo ((n+1)^2-n^2)/6^(n+1)`

`=1/6+sum_(n=1)^oo (2n+1)/6^(n+1)`

`=1/6+1/6 sum_(n=1)^oo (2n+1)/6^n`

`=1/6+1/3 sum_(n=1)^oo n/6^n+1/6 sum_(n=1)^oo 1/6^n`

`=1/6+1/3 (6/25)+1/6 (1/5)`

`=7/25`

Multiply both ends by `6/5` to get:

`sum_(n=1)^oo n^2/6^n = 42/125`

Answer 2

`sum_(k=0)^ook^2 6^(-k) =42/125`

Explanation:

`sum_(k=0)^ook^2x^(-k) = -x d/(dx)(sum_(k=0)^ook x^(-k))`

and

`sum_(k=0)^ookx^(-k) = -x d/(dx)(sum_(k=0)^oo x^(-k))`

if `abs(1/x) < 1` then

`sum_(k=0)^oo x^(-k) = 1/(1-1/x) = x/(x-1)`

Applying the transformations we have

`sum_(k=0)^ook^2x^(-k) = (x (x+1))/(x-1)^3`

Now, if `x=6` we have

`sum_(k=0)^ook^2 6^(-k) =42/125`