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) 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)^

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(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

Similar Questions

How many words can be formed by using all the letters of the word '$TRIVANDRUM$' using each letter exactly once (in $!$)?

If all the letters of the word $'GANGARAM'$ are arranged,then the number of words in which exactly two vowels are together but no two $'G'$ occur together is-

If $^nC_r = 84$,$^nC_{r-1} = 36$,and $^nC_{r+1} = 126$,then $n$ equals:

In how many different ways can the letters of the word '$DETAIL$' be arranged in such a way that the vowels occupy only the odd positions?

Let $S$ be the set of all triangles in the $xy$-plane,each having one vertex at the origin and the other two vertices lying on the coordinate axes with integral coordinates. If each triangle in $S$ has an area of $50$ sq. units,then the number of elements in the set $S$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module