The sum sum_{r=1}^{10} (r^2 + 1) times (r!) i | vedclass

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

Question

(B) Let the general term be $T_r = (r^2 + 1)r!$.We can rewrite the term as: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$.This does not simplify ea

Vedclass · Accepted Answer

$10 	imes (11!)$

Vedclass · Answer

(B) Let the general term be $T_r = (r^2 + 1)r!$.We can rewrite the term as: $T_r = (r^2 + r - r + 1)r! = (r(r+1) - (r-1))r!$.This does not simplify easily,so let's use the form: $T_r = (r^2 + r - r + 1)r! = r(r+1)r! - (r-1)r!$.Note that $r(r+1)r! = r(r+1)!$.So,$T_r = r(r+1)! - (r-1)r!$.This is a telescoping series of the form $f(r) - f(r-1)$ where $f(r) = r(r+1)!$.Summing from $r=1$ to $10$:$\sum_{r=1}^{10} (r(r+1)! - (r-1)r!) = [1(2!) - 0(1!)] + [2(3!) - 1(2!)] + [3(4!) - 2(3!)] + \dots + [10(11!) - 9(10!)]$.All intermediate terms cancel out,leaving the last term: $10(11!)$.

The sum $\sum_{r=1}^{10} (r^2 + 1) \times (r!)$ is equal to

Similar Questions

$A$ set contains $(2n + 1)$ elements. The number of subsets of the set which contain at most $n$ elements is:

The number of $5$ digit telephone numbers having at least one of their digits repeated is

$A$ candidate is required to answer $6$ out of $10$ questions,which are divided into two groups each containing $5$ questions. The candidate is not permitted to attempt more than $4$ questions from each group. In how many ways can the candidate make their choice?

The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to

In how many different ways can the letters of the word '$CORPORATION$' be arranged so that the vowels always come together?

Vedclass Test Series

Exam Paper Generator

Online Exam Module