# 4 + 6 + 9 + 13 + 18 + ... - Mathematics

4 + 6 + 9 + 13 + 18 + ...

#### Solution

Let $T_n$  be the nth term and  $S_n$ be the sum of n terms of the given series.

Thus, we have:

$S_n = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_n$ ...(1)

Equation (1) can be rewritten as:

$S_n = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_n$  ...(2)

On subtracting (2) from (1), we get:

$S_n = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_n$

$S_n = 4 + 6 + 9 + 13 + 18 + . . . + T_{n - 1} + T_n$

$0 = 4 + \left[ 2 + 3 + 4 + 5 + 6 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n$

The sequence of difference between successive terms is 2, 3, 4, 5,...
We observe that it is an AP with common difference 1 and first term 2.
Now,

$4 + \left[ \frac{\left( n - 1 \right)}{2}\left\{ 4 + \left( n - 2 \right)1 \right\} \right] - T_n = 0$

$\Rightarrow 4 + \left[ \frac{\left( n - 1 \right)}{2}\left( n + 2 \right) \right] - T_n = 0$

$\Rightarrow 4 + \left[ \frac{n^2 + n}{2} - 1 \right] - T_n = 0$

$\Rightarrow \left[ \frac{n^2}{2} + \frac{n}{2} + 3 \right] = T_n$

$\because S_n = \sum^n_{k = 1} T_k$

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

$= \frac{1}{2} \sum^n_{k = 1} k^2 + \frac{1}{2} \sum^n_{k = 1} k + \sum^n_{k = 1} 3$

$= \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{2 \times 6} + \frac{n\left( n + 1 \right)}{2 \times 2} + 3n$

$= n\left( \frac{2 n^2 + 3n + 1 + 3n + 3 + 36}{12} \right)$

$= \frac{n}{12}\left( 2 n^2 + 6n + 40 \right)$

$= \frac{n}{6}\left( n^2 + 3n + 20 \right)$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.2 | Q 7 | Page 18