Question

The nth term of the series $1+2+5+12+25+....$ is

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

Answer 1

Answer: (B)

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