# Pattern recognition for $S_n=2+5+13+35+…+n^{th}_{term}$

I want to find sum of the first $n^{th}$ term of this sqequence . $$2,5,13,35,97,275,793,...\\s_n=2+5+13+35+97+...$$

What is the closed form formula for $s_n$?

• The sequence $2,5,13,35,97,275,793, \dots$ is oeis.org/A007689 – lhf Aug 18 '17 at 11:53

If you look at number again , in your problem .It would be $$2,5,13,35,97,275,393,...\\1+1,2+3,4+9,8+27,16+81,32+243,...$$ so it would be $$2^0+3^0,2^1+3^1,2^2+3^2+,...\\\implies a_n=2^{n-1}+3^{n-1}$$
$S_n$ is sum of two geometric progression $$s_n=\sum_{k=0}^{n}(2^{k}+3^{k})=\\\sum_{k=0}^{n}(2^{k})+\sum_{k=0}^{n}(3^{k})=\\1.\cdot\frac{2^{n}-1}{2-1}+1.\cdot\frac{3^{n}-1}{3-1}$$
Alternatively, the sequence is: $$2,3\cdot 2-1,3\cdot 5-2,3\cdot 13-4,3\cdot 35-8,...$$ It is the recurrence relation: $$a_n=3a_{n-1}-2^{n-2},a_1=2.$$ Divide it by $2^{n}$: $$\frac{a_n}{2^{n}}=\frac{3a_{n-1}}{2^{n}}-\frac14.$$ Denote: $b_n=\frac{a_n}{2^n}$ to get: $$b_n=\frac{3}{2}b_{n-1}-\frac{1}{4},b_1=1.$$ Solution is: $$b_n=\frac{1}{3}\left(\frac{3}{2}\right)^n+\frac12.$$ Hence: $$a_n=2^nb_n=2^{n-1}+3^{n-1}.$$ Now the sum $S_n$ is calculated in the same way as in previous solution.