(a+1/a)^2+(a^2+1/a^2)^2+......+(a^n+1/a^n)^2=A?

Question

give a formula for finding A.

Answer 1

For a^2 not equal to 1,

A = [a^2(a^2n - 1) + (1 - 1/(a^(2n))] / (a^2 - 1) + 2n.

For a = 1, A = 4n.

For a = - 1, the result depends on ' m ' where n is (2m) or (2m + 1), (starting only with m = 0 and n = 1, and subsequently for both numbers of this form with m = 1, 2, ...). Then A = 4m.

How to find these three kinds of result:

1.) For a^2 not equal to 1: The rth term of the original series is (a^r + 1/a^r)^2 = a^(2r) + 2 +1/a^(2r). So the entire series can be considered as the sum of three separate series with individual rth terms:

a^(2r), 1/a^(2r), and 2.

The result given in the first line is the simplified sum of the first two separate series with general terms just given, plus another sum which is simply n times 2.

2.) For a = 1, the original series simply becomes n terms, each of size 2^2 = 4, so the sum is 4n.

3,) For a = - 1, all the odd terms become zero. Only the even terms contribute an amount (4, again) to the final sum, and the latter only increases when the series ends with an even-numbered term again. The resulting sum is

0, 4, 4, 8, 8, 12, 12,..., for which the ' m ' values are
0, 1, 1, 2, 2, 3, 3, ... .

Hence in this last case the sum A is 4m at any stage.

Live long and prosper.