Verified Solution • 2025–26 Session moderate

QJEEM21Ma14c4: JEE Main 2021 Mathematics - Class 11 JEEM21Ma14c4 Solution: If , then the value of is equal to ___________.

If ∑r=110r!(r3+6r2+2r+5)=α(11!)\sum\limits_{r = 1}^{10} {r!({r^3} + 6{r^2} + 2r + 5) = \alpha (11!)} , then the value of α\alpha is equal to ___________.

Step-by-Step Verified Solution

∑r=110r![(r+1)(r+2)(r+3)−9(r+1)+8]\sum\limits_{r = 1}^{10} {r![(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8]}

=∑r=110[{(r+3)!−(r+1)!}−8{(r+1)!−r!}]= \sum\limits_{r = 1}^{10} {[\{ (r + 3)! - (r + 1)!\} - 8\{ (r + 1)! - r!\} ]}

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

=(12 . 13+12−8) . 11!−8+8=(160)(11!)= (12\,.\,13 + 12 - 8)\,.\,11! - 8 + 8 = (160)(11!)

Therefore, α=160\alpha = 160

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