Question

Find the function of the series sum (x^2 +1)^2k

Answer 1

we do this in steps, and of course the answer will only be valid for \(|x^2+1|^2<1\)

Answer 2

trick is to make it look like the geometric series
\[\sum x^k=\frac{1}{1-x}\] or rather to use that as a starting point

Answer 3

what is the geon. series...

Answer 4

then
\[\sum x^{2k}=\frac{1}{1-x^2}\] by replacing \(x\) by \(x^2\)

Answer 5

hold the phone
is this line ok?
\[\sum x^k=\frac{1}{1-x}\]

Answer 6

because that is our starting point. if that is not clear, then we cannot continue
if that is clear, then it is algebra from here on in

Answer 7

but keep i mind at this is \[1\div1+x ^{2}\] not the 1/1-x...

Answer 8

we have this \(\sum (x^2+1)^2\) right?

Answer 9

\[\sum_{}^{} (x ^{2}+1)^{2k}\]

Answer 10

so we are going to make some replacements in this identity
\[\sum x^k=\frac{1}{1-x}\]
first we find
\[\sum x^{2k}=\frac{1}{1-x^2}\] by replacing \(x\) by \(x^2\)
then we replace \(x\) in the above by \((x^2+1)\) to get
\[\sum (x^2+1)^{2k}=\frac{1}{1-(x^2+1)^2}\]

Answer 11

then maybe a little alegra to clean up the denominator and write
\[\sum (x^2+1)^{2k}=\frac{1}{x^2(x^2+2)}\]

Answer 12

*algebra