sumlimits_{r = 0}^{100} {(r^2 + 4r + 4)(r + 1 | vedclass

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(B) The given expression is $\sum\limits_{r = 0}^{100} {(r + 2)^2 (r + 1)!}$.We can write $(r + 2)^2 (r + 1)!$ as $(r + 2) \cdot (r + 2)!$.To use the

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$(103)! - 2$

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(B) The given expression is $\sum\limits_{r = 0}^{100} {(r + 2)^2 (r + 1)!}$.We can write $(r + 2)^2 (r + 1)!$ as $(r + 2) \cdot (r + 2)!$.To use the method of differences,we express $(r + 2) \cdot (r + 2)!$ as $(r + 3 - 1)(r + 2)! = (r + 3)! - (r + 2)!$.Thus,the sum is $\sum\limits_{r = 0}^{100} {[(r + 3)! - (r + 2)!]}$.This is a telescoping sum:$[(3! - 2!) + (4! - 3!) + (5! - 4!) + \dots + (103! - 102!)]$.All intermediate terms cancel out,leaving $(103! - 2!)$.Since $2! = 2$,the sum is $(103! - 2)$.

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

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