Question

If $S_n$ and $T_n$ represent the sum of $n$ terms and the $n^{\text{th}}$ term respectively, of the series $1+4+10+20+35+\cdots$, then the value of $\frac{19T_{20}}{S_{19}}$ is

Answer 1

$S_n=1+4+10+20+35+\cdots +T_n{S}_n=1+4+10+20+35+\cdots +T_{n-1}+T_n$

$T_n=1+3+6+10+15+\cdots +t_n$, where $t_n=T_n-T_{n-1}$
$T_n=1+3+6+10+15+\cdots +t_{n-1}+t_n$

$\Rightarrow t_n=1+2+3+4+\cdots \text{upto}n\text{terms}\Rightarrow t_n=\frac{1}{2}n(n+1)\Rightarrow T_n=\sum _{r=0}^n{t}_r=\frac{1}{2}\sum _{r=0}^{n}r(r+1)\Rightarrow T_n=\frac{1}{2}\left[\frac{n(n+1)(n+2)}{3}\right]\Rightarrow T_n=\frac{1}{6}n(n+1)(n+2)$

$S_n=\sum _{r=0}^n{T}_r\Rightarrow S_n=\frac{1}{6}\sum _{r=0}^{n}r(r+1)(r+2)\Rightarrow S_n=\frac{1}{24}n(n+1)(n+2)(n+3)$

So, $\frac{n{T}_{n+1}}{S_n}=\frac{\frac{1}{6}n(n+1)\left[\right(n+1)+1]\left[\right(n+1)+2]}{\frac{1}{24}n(n+1)(n+2)(n+3)}$
$\therefore \frac{19T_{20}}{S_{19}}=4$