GATE Chemical 2020 | GA | Question: 5

Question

The difference between the sum of the first $2n$ natural numbers and the sum of the first $n$ odd natural numbers is ______

• A. $n^2-n$
• B. $n^2+n$
• C. $2n^2-n$
• D. $2n^2+n$

Comments

Put n=1, match option.

Answer 1

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 $(B).$

Answer 2

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 $(B).$

Answer 3

$\therefore$ The answer is option $\LARGE B$

Answer 4

$\therefore$ The answer is option $\LARGE B$

Answer 5

Just simply take n as 5 therefore sum of first 2n natural number will be (1+2+3+4+5+6+7+8+9+10) = 55 and first n natural numbers (1+3+5+7+9)= 25 therefore difference is 30 susbtitiute the values in the option you will get "B" as correct option

Answer 6

Just simply take n as 5 therefore sum of first 2n natural number will be (1+2+3+4+5+6+7+8+9+10) = 55 and first n natural numbers (1+3+5+7+9)= 25 therefore difference is 30 susbtitiute the values in the option you will get "B" as correct option

Answer 7

See Sum of first 2n natural numbers is : 1+2+3+......2n  ------eq 1

Sum of first n odd nombers is: 1+3+5+7.....(2n-1) --------eq2

If you subtract 1 from 2 you will get sum of first n even numbers : 2+4+6+......2n

Sum = n (2+2n) /2 = n*(1+n) = n+n^2 option b

Answer 8

See Sum of first 2n natural numbers is : 1+2+3+......2n  ------eq 1

Sum of first n odd nombers is: 1+3+5+7.....(2n-1) --------eq2

If you subtract 1 from 2 you will get sum of first n even numbers : 2+4+6+......2n

Sum = n (2+2n) /2 = n*(1+n) = n+n^2 option b

Answer 9

Sum of n natural no. = (n(n+1)/2)    ------> saying sum of even no.   and here n=2n
Sum of n odd no. = n^2

therefore when we solve this : (2n(2n+1))/2 - n^2
                                               = n^2 + n

Answer 10

Sum of n natural no. = (n(n+1)/2)    ------> saying sum of even no.   and here n=2n
Sum of n odd no. = n^2

therefore when we solve this : (2n(2n+1))/2 - n^2
                                               = n^2 + n