JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 15)
If $$\sum\limits_{r = 1}^{10} {r!({r^3} + 6{r^2} + 2r + 5) = \alpha (11!)} $$, then the value of $$\alpha$$ is equal to ___________.
Answer
160
Explanation
$$\sum\limits_{r = 1}^{10} {r![(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8]} $$
$$ = \sum\limits_{r = 1}^{10} {[\{ (r + 3)! - (r + 1)!\} - 8\{ (r + 1)! - r!\} ]} $$
$$ = (13! + 12! - 2! - 3!) - 8(11! - 1)$$
$$ = (12\,.\,13 + 12 - 8)\,.\,11! - 8 + 8 = (160)(11!)$$
Therefore, $$\alpha = 160$$
Comments (0)
