Find sum of $n$ terms of the series $12+14+24+58+164+\cdots$

Question

Find sum of $n$ terms: $12+14+24+58+164+\cdots$ I have tried my best but could not proceed

Answer 1

$12 + 14 + 24 + 58 + 164 + \ldots = 2 ( 6 + 7 + 12 + 29 + 82 + \ldots)$

Is the next term $486$?

You should see the second difference multiply by 3 in each succession (i.e. the difference between the terms, factored by $2$, being $1, 5, 17, 53, ...$).

Did you notice that?

Answer 2

You can do this by grouping terms and using the geometric sum formula: $$2[\{6+(1-1)\}+\{6+(3-2)\}+\{6+(9-3)\}+\{6+(27-4)\}+\dots]$$ $$=2[6n+(1+3+9+27+...)-(1+2+3+4+...)]$$ $$=2\bigg[6n+\frac{3^n-1}{3-1}-\frac{n(n+1)}{2}\bigg]$$ $$=12n+3^n-1-n^2-n$$ $$=11n+3^n-1-n^2$$