Question

The $n$th term of the series $1+2+5+12+25+\cdots$ is

• A. $(n-1)(n-2)$
• B. $\frac{1}{3}n(n-1)(n-2)+n$
• C. $n$
• D. None of these

Answer 1

The correct option is B $\frac{1}{3}n(n-1)(n-2)+n$

Let $n$th term of the series is $t_n$ and sum is $S$.
Then, $S=1+2+5+12+25+46+\cdots +t_n$
$S=0+1+2+5+12+25+\cdots +t_{n-1}+t_n$
On subtracting, we get
$0=1+1+3+7+13+21+\cdots +(t_n-t_{n-1})-t_n$
$\therefore t_n=1+\{1+3+7+13+21+\cdots +upto(n-1)\}$
Let $(n-1)$th term and sum of the series $1+3+7+13+21+\cdots$ are $t_{n-1}$ and $S^{{}^{\prime }}$, respectively. Then,
$S^{{}^{\prime }}=1+3+7+13+21+\cdots +t_{n-1}$
$S^{{}^{\prime }}=0+1+3+7+13+\cdots +t_{n-2}+t_{n-1}$
On subtracting, we get
$0=1+2+4+6+8+\cdots +(t_{n-1}-t_{n-2})-t_{n-1}$
$\therefore t_{n-1}=1+2\{1+2+3+4+\cdots upto(n-2)\}$
$=1+2\cdot \frac{1}{2}(n-2)(n-1)=n^2-3n+3$
$\Rightarrow t_n={(n+1)}^2-3(n+1)+3$
$=n^2-n+1$
$\therefore t_n=1+\{1+3+7+13+\cdots upto(n-1)\}$
$=1+\sum _{n=1}^{n-1}(n^2-n+1)$
$=1+\sum _{n=1}^{n-1}{n}^2-\sum _{n=1}^{n-1}n+\sum _{n=1}^{n-1}1$
$=1+\frac{1}{6}n(n-1)(2n-1)-\frac{1}{2}n(n-1)+(n-1)$
$=\frac{1}{3}n(n-1)(n-2)+n$
Hence, $t_n=\frac{1}{3}n(n-1)(n-2)+n$