If ∑r=110r!r^3+6r^2+2r+5=α11!, then the value of α is equal to ___ .

Correct answer: 160. ∑r=110r!r+1r+2r+3-9r+1+8 =∑r=110(r+3)!-(r+1)!-8r+1!-r! =(13!+12!-2!-3!)-8(11!-1) =(12.13+12-8)·11!-8+8 =(160)(11)! Hence α=160

Q60 / 90 · JEE Main 2021 (18 Mar Shift 2) · Mathematics

If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to ___ .

Correct answer: 160

$\sum_{r = 1}^{10} r! \{(r + 1) (r + 2) (r + 3) - 9 (r + 1) + 8\}$

$= \sum_{r = 1}^{10} [\{(r + 3)! - (r + 1)!\} - 8 \{(r + 1)! - r!\}]$

$= (13! + 12! - 2! - 3!) - 8 (11! - 1)$

$= (12.13 + 12 - 8) \cdot 11! - 8 + 8$

$= (160) (11)!$

Hence $α = 160$

Hence, the correct answer is 160.