sumlimits_{r = 0}^{100} {({r^2} + 4r + 4)(r + | vedclass

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(A) We are given the sum $S = \sum\limits_{r = 0}^{100} {(r^2 + 4r + 4)(r + 1)!}$.
Note that $r^2 + 4r + 4 = (r + 2)^2$.
So,$S = \sum\limits_{r = 0}

Vedclass · Accepted Answer

$(103)! - 2$

Vedclass · Answer

(A) We are given the sum $S = \sum\limits_{r = 0}^{100} {(r^2 + 4r + 4)(r + 1)!}$.
Note that $r^2 + 4r + 4 = (r + 2)^2$.
So,$S = \sum\limits_{r = 0}^{100} {(r + 2)^2 (r + 1)!}$.
We can rewrite $(r + 2)^2$ as $(r + 2)(r + 2) = (r + 2)(r + 3 - 1) = (r + 2)(r + 3) - (r + 2)$.
Thus,$S = \sum\limits_{r = 0}^{100} {(r + 2)(r + 3)(r + 1)! - (r + 2)(r + 1)!}$.
Since $(r + 3)(r + 2)(r + 1)! = (r + 3)!$ and $(r + 2)(r + 1)! = (r + 2)!$,we have:
$S = \sum\limits_{r = 0}^{100} {(r + 3)! - (r + 2)!}$.
This is a telescoping sum:
$S = [(3! - 2!) + (4! - 3!) + (5! - 4!) + ... + (103! - 102!)]$.
All intermediate terms cancel out,leaving $S = 103! - 2! = 103! - 2$.

$\sum\limits_{r = 0}^{100} {({r^2} + 4r + 4)(r + 1)!}$ is equal to :-

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