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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given expression is $1 + \sum\limits_{r = 0}^{22} {\left( {r^2 + 2r + 1} \right)r!} = 1 + \sum\limits_{r = 0}^{22} {\left( {r + 1} \right)^2 r

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The given expression is $1 + \sum\limits_{r = 0}^{22} {\left( {r^2 + 2r + 1} ight)r!} = 1 + \sum\limits_{r = 0}^{22} {\left( {r + 1} ight)^2 r!}$.We can rewrite the term inside the summation as:$(r+1)^2 r! = (r+1)(r+1)r! = (r+1)(r+1)! = (r+2-1)(r+1)! = (r+2)! - (r+1)!$.Substituting this into the sum:$1 + \sum\limits_{r = 0}^{22} {\left( {(r + 2)! - (r + 1)!} ight)}$.This is a telescoping sum:$1 + [(2! - 1!) + (3! - 2!) + \dots + (24! - 23!)]$.$1 + 24! - 1! = 1 + 24! - 1 = 24!$.Given $24! = k!$,we have $k = 24$.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

Similar Questions

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ terms is

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to

If $t_n = \frac{1}{4}(n+2)(n+3)$ for $n = 1, 2, 3, \ldots$, then $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}}$ is equal to

The sum of the first $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \dots$ is $.........$.

For positive integers $n$,if $4 a_n = (n^2 + 5n + 6)$ and $S_n = \sum_{k=1}^n \left(\frac{1}{a_k}\right)$,then the value of $507 S_{2025}$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module