If -r-110r-r3-6r2-2r-5-11- then the value of - is equal to

∑_r=1^10r!(r^3+6r^2+2r+5) =∑_r=1^10 nbsp;r![(r+1)(r+2)(r+3) 9r 1] =∑_r=1^10[(r+3)! 9r· r! r!] =∑_r=1^10[(r+3)! 9(r+1 1)·r! r!] =∑_r=1^10[(r+3)! 9(r+1)!+8r ...

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Byju's Answer

Standard XII

Mathematics

Vn Method

If ∑r=110r!r3...

Question

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

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Solution

$10 \sum r = 1 r! (r^{3} + 6 r^{2} + 2 r + 5) = 10 \sum r = 1 r! [(r + 1) (r + 2) (r + 3) - 9 r - 1] = 10 \sum r = 1 [(r + 3)! - 9 r \cdot r! - r!] = 10 \sum r = 1 [(r + 3)! - 9 (r + 1 - 1) \cdot r! - r!] = 10 \sum r = 1 [(r + 3)! - 9 (r + 1)! + 8 r!] = 10 \sum r = 1 [[(r + 3)! - (r + 1)!] - 8 [(r + 1)! - r!]] = [13! + 12! - 3! - 2!] - 8 [11! - 1!] = 11! [168 - 8] - 3! - 2! + 8 = (160) 11! ∴ α = 160$

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If 10∑r=1r!(r3+6r2+2r+5)=α(11!), then the value of α is equal to

10∑r=1r!(r3+6r2+2r+5)=10∑r=1r![(r+1)(r+2)(r+3)−9r−1]=10∑r=1[(r+3)!−9r⋅r!−r!]=10∑r=1[(r+3)!−9(r+1−1)⋅r!−r!]=10∑r=1[(r+3)!−9(r+1)!+8r!]=10∑r=1[[(r+3)!−(r+1)!]−8[(r+1)!−r!]]=[13!+12!−3!−2!]−8[11!−1!]=11![168−8]−3!−2!+8=(160)11!∴α=160

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