Question

Prove that the sum of the infinite series `(1*3)/2 + (3*5)/(2^2) + (5*7)/(2^3) + (7*9)/(2^4) ........ oo = 23`?

Answer 1

The general term is

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

Which can be rewritten as

`sum_(n = 1)^oo (4n^2 - 1)/2^n`

Which in turn can be written as

`sum_(n = 1)^oo (4n^2)/2^n - 1/2^n`

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

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

We know this first sequence will converge. This is because `2^(2 - n)` will approch `0` as `x` approaches `oo` and so the series converges.

According to wolfram alpha, the sum is `24`. I haven't figured out how to get that yet but I will be back once I figure it out.

The second series is just your run of the mill geometric series.

`s_oo = (1/2)/(1 - 1/2)`

`s_oo = (1/2)(2)`

`s_oo = 1`

So the sum of the entire sequence will be `24 - 1 = 23`, as required.

Hopefully this helps!

Answer 2

See below.

Explanation:

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

Here

`sum_(k=1)^oo 1/2^k = -1/(1/2-1)-1=1` and

`sum_(k=1)^oo (2k)^2/2^k = 4sum_(k=1)^oo k^2/2^k` and now assuming `abs(x) < 1` and considering

`sum_(k=1)^oo k^2 x^k = xd/(dx)sum_(k=1)^oo k x^k = xd/(dx)(x d/(dx)sum_(k=1)^oo x^k) = xd/(dx)(x d/(dx)(1/(1-x)-1)) = x(1/(1 - x)^2 + (2 x)/(1 - x)^3)`

then, making `x = 1/2 rArr sum_(k=1)^oo k^2 x^k = 6` and then

`sum_(k=1)^oo ((2k-1)(2k+1))/2^k =4 xx 6-1=23`