Question

$T_n=n!(n^2+1)$ and $S_n=T_1+T_2+T_3+.....T_n$ let $\frac{T_{10}}{S_{10}}=\frac{a}{b}$ where a, b are relatively prime natural numbers, then value of b-a is

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is B $9$

$\begin{array}{cc}{T}_n& =n!(n^2+1)\\ T_{10}& =10!({10}^2+1)\\ & =10!\left(101\right)\end{array}$

$\begin{array}{cc}{S}_{10}& =\sum _{r=1}^{r=10}{T}_r=\sum _{r=1}^{r=10}(r^2+1)\cdot r!\\ & =\sum _{r=1}^{r=10}\left[\right(r+1{)}^2-2r]\cdot r!\\ & =\sum _{r=1}^{r=10}\left[\right(r+1\left)\right(r+1)r!-(2r)\cdot r!]\\ & =\sum _{r=1}^{r=10}[r\cdot (r+1)!+(r+1)!-2r\cdot (r)!]\end{array}$

$S_{10}={\sum }_{r=1}^{r=10}\left[\right(r+1)\cdot (r+1)!-2r\cdot (r)!]$

$S_{10}=$

$(2\cdot 2!-2\cdot 1!)$

$+(3\cdot 3!-4\cdot 2!)$

$+(4\cdot 4!-6\cdot 3!)$

$+(5\cdot 5!--8\cdot 4!)$

$+(6\cdot 6!-10\cdot 5!)$

$+(7\cdot 7!-12\cdot 6!)$

$+(8\cdot 8!--14\cdot 7!)$

$+(10\cdot 10!--18\cdot 9!)$

$(11\cdot 11!-20\cdot 10!)$

$\begin{array}{cc}{S}_{10}=& 11\cdot 11!-2\cdot 1!-2\cdot 2!-3\cdot 3!-4\cdot 4!-5\cdot 5!\\ & -6\cdot 6!-7\cdot 7!-8\cdot 8!-9\cdot 9!-10\cdot 10!\end{array}$

$S_{10}=11\cdot 11!-1\cdot 1!-1\cdot 1!-2\cdot 2!-\dots \dots -10\cdot 10!$

$S_{10}=11\cdot 11!-1-{\sum }_{r=1}^{r=10}r\cdot \left(r\right)!$

${\sum }_{r=1}^{10}r\cdot r!={\sum }_{r=1}^{10}(r+1-1)\cdot r!$

$\begin{array}{cc}=& \sum _{r=1}^{10}\left[\right(r+1)!-r!]\\ =& [2!-1!]\\ & +[3!-2!]\\ & +[4!-3!]\\ & +⋮\\ & =[11!-10!]\end{array}$

$\begin{array}{cc}\sum _{r=1}^{10}r\cdot r!& =11!-1!\\ \therefore S_{10}& =11\cdot 11!-1-(11!-1)\\ & =11\cdot 11!-1+1-11!\\ S_{10}& =10\cdot 11!\\ \frac{T_{10}}{S_{10}}& =\frac{10!\cdot \left(101\right)}{11!\left(10\right)}=\frac{a}{b}\end{array}$

$\Rightarrow \frac{a}{b}=\frac{101}{110}$

$b-a=110-101=9$

$\therefore$ Answer: option $\left(B\right)$.