frac{1}{2 cdot 5} + frac{1}{5 cdot 8} + frac{ | vedclass

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

Question

(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.F

Vedclass · Accepted Answer

$\frac{4}{25}$

Vedclass · Answer

(A) The general term of the series is $T_n = \frac{1}{(3n-1)(3n+2)}$.We can write $T_n = \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.For $n=16$,the last term is $\frac{1}{(3(16)-1)(3(16)+2)} = \frac{1}{47 \cdot 50}$.The sum $S_{16} = \sum_{n=1}^{16} \frac{1}{3} \left( \frac{1}{3n-1} - \frac{1}{3n+2} \right)$.$S_{16} = \frac{1}{3} \left[ (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{8}) + \ldots + (\frac{1}{47} - \frac{1}{50}) \right]$.$S_{16} = \frac{1}{3} \left[ \frac{1}{2} - \frac{1}{50} \right] = \frac{1}{3} \left[ \frac{25-1}{50} \right] = \frac{1}{3} \left[ \frac{24}{50} \right] = \frac{8}{50} = \frac{4}{25}$.

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots$ to $16$ terms $=$

Similar Questions

If $|x| < 1$,what is the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$?

Find the sum to $n$ terms of the series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots$

The mean of the squares of the first $n$ natural numbers is

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ up to $8$ terms, is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module