Question

${\displaystyle \sum _{r=1}^{10}}r!\left(r^3+6r^2+2r+5\right)=\alpha \left(11!\right)$. What is the value of $\alpha$?

Answer 1

Step 1 : Finding the $r^{th}$ term

Given ${\displaystyle \sum _{r=1}^{10}}r!\left(r^3+6r^2+2r+5\right)=\alpha \left(11!\right)$.

So, the $r^{th}$ term of the given series will be :

$$\begin{gathered} T_r=r!\left(r^3+6r^2+2r+5\right) \\ =r!\left(r^3+6r^2+11r+6-9r-1\right) \\ =r!\left(r^3+r^2+5r^2+5r+6r+6-9r-1\right) \\ =r!\left[r^2\left(r+1\right)+5r\left(r+1\right)+6\left(r+1\right)-9r-1\right] \\ =r!\left[\left(r+1\right)\left(r^2+5r+6\right)-9r-1\right] \\ =r!\left[\left(r+1\right)\left(r+2\right)\left(r+3\right)-9r-1\right] \\ =r!\left(r+1\right)\left(r+2\right)\left(r+3\right)-9rr!-r!\left[\because r!\left(r+1\right)\left(r+2\right)\left(r+3\right)=\left(r+3\right)!\right] \\ =\left(r+3\right)!-9\left(r+1-1\right)r!-r! \\ =\left(r+3\right)!-9\left(r+1\right)r!+9r!-r! \\ =\left(r+3\right)!-9\left(r+1\right)!+8r!\left[\because \left(r+1\right)r!=\left(r+1\right)!\right] \\ =\left(r+3\right)!-\left(r+1\right)!-8\left[\left(r+1\right)!-r!\right] \end{gathered}$$

Thus $T_r=\left(r+3\right)!-\left(r+1\right)!-8\left[\left(r+1\right)!-r!\right]$.

Step 2 : Finding the value of $\alpha$

Using the above expression for $T_r$, we get

$$\begin{gathered} \sum _{r=1}^{10}{T}_r=\sum _{r=1}^{10}\left[\left(r+3\right)!-\left(r+1\right)!-8\left\{\left(r+1\right)!-r!\right\}\right] \\ =\left[4!-2!-8\left\{2!-1!\right\}\right]+\left[5!-3!-8\left\{3!-2!\right\}\right]+\left[6!-4!-8\left\{4!-3!\right\}\right]+\cdots \left[13!-11!-8\left\{11!-10!\right\}\right] \\ =(13!+12!-3!-2!)-8\left[11!-1!\right] \\ =13!+12!-8-8\times 11!+8 \\ =13\times 12\times 11!+12\times 11!-8\times 11! \\ =(13\times 12+12-8)\times 11! \\ =160\times 11! \end{gathered}$$

Comparing the given equation , $\sum _{r=1}^{10}{T}_r=\alpha \left(11\right)!$ and $\sum _{r=1}^{10}{T}_r=160\left(11\right)!$, we get $\alpha =160$

Hence, the value of $\alpha =160$.