Question

The ratio of the sums of first $n$ even number and $n$ odd number will be

• A. $1:n$
• B. $\left(n+1\right):1$
• C. $\left(n+1\right):n$
• D. $\left(n-1\right):1$

Answer 1

The correct option is C

$\left(n+1\right):n$

Explanation for the correct option:

Step $1$: Compute the sum of first $n$ even numbers.

Sum of first $n$even numbers is given by ,

$2+4+6+......n$

Following series is an A.P. having $a=2,d=2$ .

Since, sum of $n$ terms of A.P. is given by,

$S_{ne}=\frac{n}{2}\left[2a+(n-1)d\right]$

$$\begin{gathered} \Rightarrow S_{ne}=\frac{n}{2}\left[4+(n-1)2\right] \\ \Rightarrow S_{ne}=n\left[2+n-1\right] \\ \Rightarrow S_{ne}=n\left(1+n\right) \end{gathered}$$

Step $2$: Compute the sum of first $n$ odd numbers.

Sum of first $n$odd numbers is given by ,

$1+3+5+......n$

Following series is an A.P. having $a=1,d=2$

.$$\begin{gathered} S_{no}=\frac{n}{2}\left[2+(n-1)2\right] \\ \Rightarrow S_{no}=n\left[1+n-1\right] \\ \Rightarrow S_{no}=n^2 \end{gathered}$$

Step $3$: Compute the ratio.

$$\begin{gathered} \frac{S_{ne}}{S_{no}}=\frac{n(1+n)}{n^2} \\ \Rightarrow \frac{S_{ne}}{S_{no}}=\frac{(1+n)}{n} \end{gathered}$$

Hence option (C) is the correct option.