Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

1 1 . 4 + 1 4 . 7 + 1 7 . 10 + . . . - Mathematics

$\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .$

Solution

Let $T_n$  be the nth term of the given series.
Thus, we have:

$T_n = \frac{1}{(3n - 2) (3n + 1)}$

Now, let

$S_n$

be the sum of n terms of the given series.
Thus, we have:

$S_n = \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)\left( 3k + 1 \right)}$

$= \frac{1}{3} \sum^n_{k = 1} \left( \frac{1}{\left( 3k - 2 \right)} - \frac{1}{\left( 3k + 1 \right)} \right)$

$= \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)} - \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k + 1 \right)}$

$= \frac{1}{3}\left[ \left( 1 + \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + . . . + \frac{1}{3n - 2} \right) - \left( \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + . . . + \frac{1}{3n - 2} + \frac{1}{3n + 1} \right) \right]$

$= \frac{1}{3}\left[ 1 - \left( \frac{1}{3n + 1} \right) \right]$

$= \frac{n}{3n + 1}$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.2 | Q 9 | Page 18