The series is neither AP nor GP, so I tried doing some manipulations to get the expression for the general term.
$S = 9 + 16 + 29 + 54 + 103 + … T_n$
$S = 0 + 9 + 16 + 29 + 54 + … T_{n-1} + T_n$
Subtracting the two:
$0 = 9 + 7 + 13 + 25 + 49 + … (T_n – T_{n-1}) – T_n$
This gives:
$T_n = 9 + 7 + 13 + 25 + 49 + … (T_n – T_{n-1})$
The expression obtained for $T_n$ is neither in AP nor in GP.
I repeated the above process with $T_n$ (which gave me a GP this time) and got this expression finally:
$T_n – T_{n-1} = 7 + 6(2^{n-2} – 1) = 6(2^{n-2}) + 1$
How do I proceed from here to get an expression for the general term as well as the sum of the series?
Best Answer
From $T_n - T_{n-1} = 6(2^{n-2}) + 1$ you have $$T_n = T_0 + \sum_{k=1}^n (6(2^{n-2}) + 1) \\ = T_0 + n + 3 (2^n -1)$$ The general approach is: $$T_n = a+ n + b 2^n$$ so $$9 = T_1 = a + 1 + 2 b\\ 16 = T_2 = a + 2 + 4b $$ which gives indeed $b = 3$ and $a = 2$, hence $$T_n = 2+ n + 3 \cdot 2^n$$
The sum is then $$ S_n = \sum_{k=1}^n T_k = 2 n + \frac12 n (n+1) + 3 (2^{n+1} -1) $$