Name: Sum of First n Natural Numbers
Uploaded: 2019-07-19
Duration: 3 min
Description: What are natural numbers? Do you know how to find the sum of the first n natural numbers? Watch this video to get answers to these questions.

Question

Find the sum of the series:

$(1 \times 2 \times 3) + (2 \times 3 \times 4) + (3 \times 4 \times 5) + \dots$ to $n$ terms.

Accepted Answer

The series given is $(1 \times 2 \times 3) + (2 \times 3 \times 4) + (3 \times 4 \times 5) + \dots$ to $n$ terms.

The series can be written as, $[(1 \times (1 + 1) \times (1 + 2)), (2 \times (2 + 1) \times (2 + 2) . . . (n \times (n + 1) x (n - 2)]$

So, $n^{th}$ term of the series, $a_{n} = n (n + 1) (n + 2)$

${⇒a}_{n} = n^{3} + 3 n^{2} + 2 n$

Now, we need to find the sum of this series, $S_{n}$ .

${⇒S}_{n} = \sum_{n = 1}^{n} a_{n}$

${⇒S}_{n} = \sum_{n = 1}^{n} (n^{3} + 3 n^{2} + 2 n)$

$\Rightarrow S_{n} = \sum_{n = 1}^{n} (n^{3}) + 3 \sum_{n = 1}^{n} (n^{2}) + 2 \sum_{n = 1}^{n} (n)$

Sum of first $n$ natural numbers, $1 + 2 + 3 + \dots n$ , $\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

Sum of squares of first $n$ natural numbers, $1^{2} + 2^{2} + 3^{2} + \dots . n^{2}$ $, \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Sum of cubes of first $n$ natural numbers, $1^{3} + 2^{3} + 3^{3} + \dots \dots n^{3}$ $, \sum_{k = 1}^{n} k^{3} = {(\frac{n (n + 1)}{2})}^{2}$

${∴S}_{n} = \sum_{n = 1}^{n} (n^{3}) + 3 \sum_{n = 1}^{n} (n^{2}) + 2 \sum_{n = 1}^{n} (n)$

${⇒S}_{n} = {(\frac{n (n + 1)}{2})}^{2} + 3 (\frac{n (n + 1) (2 n + 1)}{6}) + 2 (\frac{n (n + 1)}{2})$

$= (\frac{n (n + 1)}{2}) [\frac{n (n + 1)}{2} + (2 n + 1) + 2]$

$= (\frac{n (n + 1)}{2}) [\frac{n^{2} + 5 n + 6}{2}]$

$= (\frac{n (n + 1)}{2}) [\frac{(n + 3) (n + 2)}{2}]$

So, sum of the series, $S_{n} = (\frac{n (n + 1) (n + 2) (n + 3)}{4})$ .

R S Aggarwal Solutions for Exercise 1: EXERCISE 13A

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE 13A

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