Question

If the $n^{\text{th}}$ term and the sum of $n$ terms of the series $2,12,36,80,150,252,.....$ is $T_n$ and $S_n$ repectively then

• A. $T_n=n^3+n^2$
• B. $\frac{1}{12}\left(n\right)(n+1)(n+2)(3n+2)$
• C. $T_n=n^3-n^2$
• D. $\frac{1}{12}\left(n\right)(n+1)(n+2)(3n+1)$

Answer 1

Answer: (A), (D)

$\frac{1}{12}\left(n\right)(n+1)(n+2)(3n+1)$

$\text{Let}{S}_n=2+12+36+80+150+252+......+T_n\cdots \left(1\right)S_n=2+12+36+80+150+252+....+T_{n-1}+T_n\cdots \left(2\right)\text{On subtracting equation (2) from equation (1) ,We get}{T}_n=2+10+24+44+70+102+...+t_n\cdots \left(3\right),\text{where}{t}_n=T_n-T_{n-1}{T}_n=2+10+24+44+70+102+...+t_{n-1}+t_n\cdots \left(4\right)\text{On subtracting equation (4) from equation (3) ,We get}{t}_n=2+8+14+20+26+...=\frac{n}{2}[4+(n-1\left)6\right]=n[3n-1]t_n=3n^2-n\therefore \text{General term of given series is}{T}_n=\sum t_n=\sum (3n^2-n)=\frac{n(n+1)(2n+1)}{2}-\frac{n(n+1)}{2}=n^3+n^2\text{Hence,sum of the series is}{S}_n=\sum n^3+\sum n^2=\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{6}=\frac{1}{12}\left(n\right)(n+1)(n+2)(3n+1)$

Alternate Solution:
Substitute $n=1,2,3\cdots$ in the options given
and verify it with the given series