The natural numbers are divided into groups as $(1), (2,3,4), (5,6,7,8,9), \dots $ and so on. Then, the sum of the numbers in the $15 \text{th}$ group is equal to

- $6090$
- $4941$
- $6119$
- $7471$

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1
votes

The natural numbers are divided into groups as $(1), (2,3,4), (5,6,7,8,9), \dots $ and so on. Then, the sum of the numbers in the $15 \text{th}$ group is equal to

- $6090$
- $4941$
- $6119$
- $7471$

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Best answer

3
votes

We can write the number in the $ n^{\text{th}}$ group.

- The number in $1^{\text{st}}$ group $ = (1)$
- The number in $2^{\text{nd}}$ group $ = (2, 3, 4)$
- The number in $3^{\text{rd}}$ group $ = (5, 6, 7, 8, 9)$
- $\vdots \quad \vdots \quad \vdots $
- The number in $ n^{\text{th}}$ group $ = \left( n^{2} – (2n-2), n^{2} – (2n-1), \dots, n^{2} – 1, n^{2} \right)$

The number of element in $n^{\text{th}}$ group $ = 2n – 1$

So, the sum of numbers of $n^{\text{th}}$ group

$S_{n} = n^{2} + (n^{2}-1) + (n^{2}-2) + \dots + \left( n^{2} – (2n-2) \right)$

$\Rightarrow S_{n} = (2n – 1)n^{2} – \left( 1 + 2 + 3 + \dots + (2n-2) \right)$

$\Rightarrow S_{n} = (2n-1)n^{2} + \frac{(2n-2)(2n-1)}{2}$

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

$ \Rightarrow S_{n} = (2n-1)n^{2} + 2n^{2} – n – 2n + 1$

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

$\Rightarrow S_{n} = (2n-1) (n^{2} – n + 1) $

$\Rightarrow S_{n} = 2n^{3} – 3n^{2} + 3n – 1 $

$\Rightarrow S_{n} = n^{3} + n^{3} – 3n^{2} + 3n – 1$

$\Rightarrow S_{n} = n^{3} + (n-1)^{3} \quad [\because n^{3} – 3n^{2} + 3n – 1 = (n-1)^{3}]$

$\therefore$ The sum of the numbers in the $15^{\text{th}}$ group $ = 15^{3} + (15-1)^{3}$

$ \qquad \qquad \qquad = 15^{3} + 14^{3} = 3375 + 2744= 6119.$

Correct Answer $: \text{C}$

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