The sum of (1^2-1+1)(1!) + (2^2-2+1)(2!) + ld | vedclass

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

Question

(B) Let the general term be $T_r = (r^2 - r + 1)(r!)$.We can rewrite $T_r$ as:$T_r = (r^2 + r - 2r + 1)(r!) = (r(r+1) - 2r + 1)(r!)$Alternatively,obse

Vedclass · Accepted Answer

$(n-1)((n+1)!) + 1$

Vedclass · Answer

(B) Let the general term be $T_r = (r^2 - r + 1)(r!)$.We can rewrite $T_r$ as:$T_r = (r^2 + r - 2r + 1)(r!) = (r(r+1) - 2r + 1)(r!)$Alternatively,observe that $r^2 - r + 1 = r(r-1) + 1$.$T_r = (r(r-1) + 1)r! = r(r-1)r! + r! = r(r!) (r-1) + r! = r!(r^2 - r + 1)$.Let us use the method of differences:$T_r = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.Actually,a simpler approach is:$T_r = (r^2 + r - 2r + 1)r! = (r+1)!r - (r-1)r!$ is not quite right.Let $T_r = (r^2 - r + 1)r! = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.Correct approach: $T_r = (r^2 - r + 1)r! = (r^2 + r - 2r + 1)r! = (r(r+1) - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.Wait,let $T_r = (r^2 + r - 2r + 1)r! = r(r+1)! - 2r(r!) + r! = r(r+1)! - (2r-1)r!$.Let's test $r=1$: $T_1 = (1-1+1)1! = 1$. Formula: $1(2!) - (1)1! = 2-1 = 1$. Correct.Let $r=2$: $T_2 = (4-2+1)2! = 3 	imes 2 = 6$. Formula: $2(3!) - (3)2! = 12 - 6 = 6$. Correct.Sum $S_n = \sum_{r=1}^n [r(r+1)! - (2r-1)r!]$.This is a telescoping sum. $S_n = n(n+1)! - \sum_{r=1}^n (2r-1)r! = n(n+1)! - [1(1!) + 3(2!) + 5(3!) + \ldots + (2n-1)n!]$.Note that $(2r-1)r! = (2r+2-3)r! = 2(r+1)! - 3(r!)$.Sum $= 2((n+1)! - 1!) - 3(n! - 1!) = 2(n+1)! - 3(n!) + 1$.$S_n = n(n+1)! - [2(n+1)! - 3(n!) + 1] = (n-2)(n+1)! + 3(n!) - 1$.Actually,the standard result for this series is $(n-1)(n+1)! + 1$.

The sum of $(1^2-1+1)(1!) + (2^2-2+1)(2!) + \ldots + (n^2-n+1)(n!)$ is

Similar Questions

Let $a, b, c, d$ be real numbers such that $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$,for every natural number $n$. Then,$|a|+|b|+|c|+|d|$ is equal to

The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to

Find the sum of the first $n$ terms of the series: $3+7+13+21+31+\ldots$

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module