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

There are $3$ girls in a class of $10$ students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is

If $^nC_{12} = ^nC_6$,then $^nC_2 = $

$A$ person wants to climb a $n$-step staircase using one step or two steps. Let $C_n$ denote the number of ways of climbing the $n$-step staircase. Then $C_{18} + C_{19}$ equals

$A$ library has $a$ copies of one book,$b$ copies of each of two books,$c$ copies of each of three books,and single copies of $d$ books. The total number of ways in which these books can be arranged is

In how many ways can a garland be made from exactly $10$ flowers?

Vedclass Test Series

Exam Paper Generator

Online Exam Module