Find the $n^{th}$ term and the sum to $n$ terms of the series.

Question

Find the $n^{th}$ term and the sum to $n$ terms of the series. $$5+7+13+31+85+.....$$

I couldn't get much idea regarding this. Just an observation that the difference between the term and preceding term forms a GP with a common ratio of $3$. How do I proceed further?

Answer 1

Let $S_n=5+7+3+31+85+....$

Now $$S_n=5+7+3+31+85+....+T_{n-1}+T_n$$ $$\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }(-)S_n=\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }5+7+3+31+....+T_{n-2}+T_{n-1}+T_n$$ $$\\\rule{13cm}{1pt}$$ $$\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }0=\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }5+[2+6+18+54+....+(T_n-T_{n-1})]-T_n$$ Now,$$T_n=5+[2+6+18+54+....+(T_n-T_{n-1})]-T_n$$ $$=5+\frac{2(3^{n-1}-1)}{3-1}$$ $$=5+3^{n-1}-1$$ $$=4+3^{n-1}$$

Answer 2

Using already mentioned properties the answer is: $$ a_n=4+3^{n-1},\quad S_n=4n+\frac{3^n-1}{2}. $$