If sum_{r=1}^{10} r! (r^3 + 6r^2 + 2r + 5) = | vedclass

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(C) We express the term inside the summation as: $r^3 + 6r^2 + 2r + 5 = (r+1)(r+2)(r+3) - 9(r+1) + 8$. Thus,the sum becomes: $\sum_{r=1}^{10} [r!(r+1)

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(C) We express the term inside the summation as: $r^3 + 6r^2 + 2r + 5 = (r+1)(r+2)(r+3) - 9(r+1) + 8$. Thus,the sum becomes: $\sum_{r=1}^{10} [r!(r+1)(r+2)(r+3) - 9r!(r+1) + 8r!]$. Using the property $(r+k)! = r!(r+1)...(r+k)$,we rewrite the terms: $= \sum_{r=1}^{10} [(r+3)! - 9(r+1)! + 8r!]$. $= \sum_{r=1}^{10} [(r+3)! - (r+1)! - 8((r+1)! - r!)]$. Expanding the summation: $= [(4! - 2!) + (5! - 3!) + ... + (13! - 11!)] - 8[(2! - 1!) + (3! - 2!) + ... + (11! - 10!)]$. $= (13! + 12! - 2! - 3!) - 8(11! - 1!)$. $= (13 	imes 12 	imes 11! + 12 	imes 11! - 2 - 6) - 8(11!) + 8$. $= (156 + 12 - 8) 	imes 11! - 8 + 8$. $= 160 	imes 11!$. Comparing with $\alpha(11!)$,we get $\alpha = 160$.

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

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