The sum of the series frac{3}{{1! + 2! + 3!}} | vedclass

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

Question

(C) The general term of the series is $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$ for $n = 1, 2, \dots, 2006$.$T_n = \frac{n+2}{n! [1 + (n+1) + (n+2)(n+1

Vedclass · Accepted Answer

$\frac{{\left( {2008} \right)! - 2}}{{2.\left( {2008} \right)!}}$

Vedclass · Answer

(C) The general term of the series is $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$ for $n = 1, 2, \dots, 2006$.$T_n = \frac{n+2}{n! [1 + (n+1) + (n+2)(n+1)]}$$T_n = \frac{n+2}{n! [n+2 + (n+2)(n+1)]} = \frac{n+2}{n! (n+2) [1 + n + 1]} = \frac{n+2}{n! (n+2) (n+2)} = \frac{1}{n! (n+2)}$This does not simplify to a simple telescoping series directly. Let us re-evaluate the original series term: $T_n = \frac{n+2}{n! + (n+1)! + (n+2)!}$.For $n=1$,$T_1 = \frac{3}{1!+2!+3!} = \frac{3}{1+2+6} = \frac{3}{9} = \frac{1}{3}$.Using the telescoping form $T_n = \frac{1}{n!} - \frac{1}{(n+2)!}$ is incorrect. Let us use $T_n = \frac{n+2}{n! (1 + n+1 + (n+2)(n+1))} = \frac{n+2}{n! (n+2)^2} = \frac{1}{n!(n+2)}$.Actually,the standard approach for this series is $T_n = \frac{n+2}{n!(1 + (n+1) + (n+1)(n+2))} = \frac{n+2}{n!(n+2)^2} = \frac{1}{n!(n+2)}$.Summing this from $n=1$ to $2006$ gives $\sum_{n=1}^{2006} \frac{1}{n!(n+2)} = \sum_{n=1}^{2006} \frac{n+1}{(n+2)!} = \sum_{n=1}^{2006} \left( \frac{n+2-1}{(n+2)!} \right) = \sum_{n=1}^{2006} \left( \frac{1}{(n+1)!} - \frac{1}{(n+2)!} \right)$.This is a telescoping sum: $(\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} - \frac{1}{4!}) + \dots + (\frac{1}{2007!} - \frac{1}{2008!}) = \frac{1}{2} - \frac{1}{2008!} = \frac{2008! - 2}{2 \cdot 2008!}$.

The sum of the series $\frac{3}{{1! + 2! + 3!}} + \frac{4}{{2! + 3! + 4!}} + \frac{5}{{3! + 4! + 5!}} + ...... + \frac{{2008}}{{\left( {2006} \right)! + \left( {2007} \right)! + \left( {2008} \right)!}}$ is equal to

Similar Questions

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

Find the sum to $n$ terms of the series: $\frac{3}{1 \cdot 2} \cdot \frac{1}{2} + \frac{4}{2 \cdot 3} \cdot \left( \frac{1}{2} \right)^2 + \frac{5}{3 \cdot 4} \cdot \left( \frac{1}{2} \right)^3 + \dots$

If ${x_1}, {x_2}, {x_3}, \dots, {x_n}$ are in $A.P.$ whose common difference is $\alpha$,then the value of $\sin \alpha (\sec {x_1} \sec {x_2} + \sec {x_2} \sec {x_3} + \dots + \sec {x_{n-1}} \sec {x_n}) = $

If $t_{n} = \frac{1}{4}(n+2)(n+3)$,$n \in N$,then which one of the following is true?
Assertion $(A)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{2003}} = \frac{2003}{3009}$
Reason $(R)$ : $\frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_{n}} = \frac{4n}{3(n+3)}$

The sum of the series $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ up to $(n + 1)$ terms is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module