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?

2 Answers
Aug 27, 2017

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!

Aug 27, 2017

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