1 + sumlimits_{r = 0}^{22} {left{ {rleft( {r | vedclass

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(C) Given expression: $1 + \sum\limits_{r = 0}^{22} {\left( {r^2 + 2r + 1} \right)r!} = 1 + \sum\limits_{r = 0}^{22} {(r + 1)^2 r!}$.Note that $(r+1)^

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$8$

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(C) Given expression: $1 + \sum\limits_{r = 0}^{22} {\left( {r^2 + 2r + 1} ight)r!} = 1 + \sum\limits_{r = 0}^{22} {(r + 1)^2 r!}$.Note that $(r+1)^2 r! = (r+1)(r+1)! = (r+2-1)(r+1)! = (r+2)! - (r+1)!$.Substituting this into the sum:$1 + \sum\limits_{r = 0}^{22} {\left( {(r + 2)! - (r + 1)!} ight)}$.This is a telescoping sum:$1 + [(2! - 1!) + (3! - 2!) + \dots + (24! - 23!)]$.$1 + 24! - 1! = 1 + 24! - 1 = 24!$.So,$k! = 24!$,which implies $k = 24$.The prime factorization of $24$ is $2^3 	imes 3^1$.The number of divisors is $(3 + 1)(1 + 1) = 4 	imes 2 = 8$.

$1 + \sum\limits_{r = 0}^{22} {\left\{ {r\left( {r + 2} \right) + 1} \right\}} \cdot r! = k!$,then the number of divisors of $k$ is

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