Question

Prove that the $n$th term of the series $1+2+5+12+25+46+....$ is $T_n=\frac{n}{3}(n^3-3n+5)$

Answer 1

Let the sum of the series be $S_n$ and $n$th term of the series be $T_n$
Then $S_n=1+2+5+12+25+46+.....+T_{n-1}+T_n\cdots \left(1\right)$
$\therefore S_n=1+2+5+12+25+.......+T_{n-1}+T_n\cdots \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$, we get
$0=1+1+3+7+13+21+.......+(T_n-T_{n-1}-T_n)$
$\therefore T_n=1+3+7+13+21+.....+t_{n-1}+t_n\cdots \left(3\right)$
(Here $n$th term of $T_n$ is $t_n$
$\therefore T_n=1+1+3+7+13+......+t_{n-1}+t_n\cdots \left(4\right)$
Subtracting $\left(4\right)$ from $\left(3\right)$, we get
$0=1+0+2+4+6+8+.....+(t_n-t_{n-1})-t_n$
$\therefore t_n=1+2+4+6+8+......+(n-1)$ terms
$=1+(2+4+6+8+(n-2\left)terms\right)$
$=1+\frac{(n-2)}{2}\{2.2+(n-2-1\left)2\right\}$
$=1+(n-2)(2+n-3)$
$\therefore t_n=n^2-3n+3$
then $T_n=\sum t_n=\sum n^2-3\sum n+3\sum 1$
$=\frac{n(n+1)(2n+1)}{6}-\frac{3n(n+1)}{2}+\frac{3n}{1}$
$=\frac{n}{6}\{2n^2+3n+1-9n-9+18\}$
$\therefore T_n=\frac{n}{6}(2n^2-6n+10)$
$=\frac{n}{3}(n^2-3n+5)$