soujanyareddy13
asked
in Quantitative Aptitude
Nov 16, 2020
retagged
Feb 13, 2021
by Lakshman Patel RJIT

## 1 Answer

1 vote

Best answer

Let

- $S_{1} $ be the sum of the first $2n$ natural numbers and
- $S_{2} $ be the sum of the first $n$ odd numbers.

Now, $S_{1} = 1+2+3+ \dots + 2n = \dfrac{2n(2n+1)}{2} = n(2n+1)$

And, $S_{2} = 1+3+5+\dots +(2n-1)$

This is in Arithmetic Progression (each term at constant difference from previous term) and for arithmetic progression, the sum of first $n$ numbers is given by:

$S_{n}= \dfrac{n}{2}\left [2a+(n-1)d \right]\;\text{or}\;S_{n} = \dfrac{n}{2}\left[a+l\right]$

where,

- $n =$ number of digits in the series
- $a =$ First term of an A.P
- $d= $ Common difference in an A.P
- $l=$ last term of an A.P.

Therefore, $S_{2} = \dfrac{n}{2} [1 + (2n-1)] = \dfrac{n}{2}[2n] = n^{2}.$

$\therefore S_{1} – S_{2} = 2n^{2} + n – n^{2} = n^{2} + n.$

$\textbf{Shortcut:}$ We can take the values of $n = 1,2,3,\dots $, and get the answer.

So, the correct answer is $(B).$