Sum the following series to `n` terms: `1+4+13+40+121+`

Explanation

Let` S_n​=1+4+13+40+121+364+....T_(n+1)​+T_n​`

Rewrite` S_n`​ again
`S_n​=1+4+13+40+121+364+......+T_(n−2)​+T_(n−1)​+T_n`
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Here
`S_n​−S_(n-1)​=1+3+9+27+81+243+(T_n​−T_(n−1)​)−T_n`
​=`1+3+32+33+34+35+(T_n​−T_(n−1)​)−T_n​`

`1+3+3^2+3^3+....+(T_n​−T_(n−1)​)` is the sum of n terms of G.P.

nth term of the series
`T_n​=(3^n−1)/(3-1)​=(3^n−1)/2`
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