Question

If $S_n,T_n$ represents the sum of $n$ terms and $n^{\text{th}}$ term respectively of the series $1+4+10+20+35+\cdots$, then $\frac{n{T}_{n+1}}{S_n}=$

Answer 1

$S_n=1+4+10+20+35+\cdots +T_n\cdots \left(1\right)S_n=1+4+10+20+35+\cdots +T_{n-1}+T_n\cdots \left(2\right)$
from $\left(1\right)-\left(2\right)$
$T_n=1+3+6+10+15+\cdots +t_n\cdots \left(3\right)(\text{where}{t}_n=T_n-T_{n-1})T_n=1+3+6+10+15+\cdots +t_{n-1}+t_n\cdots \left(4\right)$
from $\left(3\right)-\left(4\right)$
$t_n=1+2+3+4+\cdots \text{upto}n\text{terms}\therefore t_n=\frac{1}{2}n(n+1)=\frac{1}{2}(n^2+n)\therefore T_n=\sum t_n=\frac{1}{2}(\sum n^2+\sum n)=\frac{1}{2}\left(\frac{1}{6}n\right(n+1\left)\right(2n+1)+\frac{1}{2}n(n+1\left)\right)=\frac{1}{6}n(n+1)(n+2)$
$\therefore S_n=\sum T_n=\frac{1}{6}(\sum n^3+\sum 3n^2+\sum 2n)=\frac{1}{6}\left(\frac{1}{4}{n}^2\right(n+1{)}^2+\frac{1}{2}n(n+1)(2n+1)+n(n+1))=\frac{1}{24}n(n+1)[n^2+n+4n+2+4]=\frac{1}{24}n(n+1)(n+2)(n+3)$
So
$\frac{n{T}_{n+1}}{S_n}=\frac{\frac{1}{6}n(n+1)\left[\right(n+1)+1]\left[\right(n+1)+2]}{\frac{1}{24}n(n+1)(n+2)(n+3)}=4$