# I would like to find the closed form of the sequence given by a <mrow class="MJX-TeX

I would like to find the closed form of the sequence given by

This task is in the topic of differential and difference equation. I don't know how to start solving this problem and what are we looking for? (${a}_{n},{a}_{n+2}$)
I do know how to solve the following form
${a}_{n+2}=2{a}_{n+1}-{a}_{n}$
using linear algebra as well. The actual problem I encountered the obstructionist term ${2}^{n}+2$
Are there some kind of variational constant method for recursive linear sequences,?
I only now this method for linear ODE with constant coefficient.
But I believe that such method could be doable here as well. Can any one provide me with a helpful hint or answer?.
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isscacabby17
Answer: By telescoping twice we obtain the following formula
${a}_{n}={2}^{n}+n\left(n-2\right)$
Enforcing ${X}_{n}={a}_{n+1}-{a}_{n}$ yields that
${a}_{n+2}=2{a}_{n+1}-{a}_{n}+{2}^{n}+2\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left({a}_{n+2}-{a}_{n+1}\right)-\left({a}_{n+1}-{a}_{n}\right)={2}^{n}+2$
${X}_{n+1}-{X}_{n}={2}^{n}+2$
By telescopic sum we have
${X}_{n+1}-{X}_{1}=\sum _{k=1}^{n}{X}_{k+1}-{X}_{k}=\sum _{k=1}^{n}\left[{2}^{k}+2\right]=2n+{2}^{n+1}-2$
That is
${a}_{n+2}-{a}_{n+1}={a}_{2}-{a}_{1}+2n+{2}^{n+1}-2=2n+{2}^{n+1}+1$
By telescopic once more we remain with
${a}_{n+2}={a}_{2}+\sum _{k=1}^{n}\left[2k+{2}^{k+1}+1\right]=4+n\left(n+1\right)+{2}^{n+2}-4+n={2}^{n+2}+n\left(n+2\right)$
Finally
${a}_{n}={2}^{n}+n\left(n-2\right)$