Question

the ratio of the sum of first n even numbers and odd numbers is ___​

Answer 1

Answer:

n square/n square+n

Step-by-step explanation:

Answer 2

The ratio of the sum of first n even numbers and odd numbers is $\frac{(n+1)}{n}$

Solution:

To find the sum of first ‘n’ even numbers we use the formula of Arithmetic Progression (A.P)

$\text { We know sum }(S)=\frac{n}{2}[2 a+(n-1) d]$

Where ‘a’ is the first number of the series, ‘d’ is the common difference between the numbers in the series and ‘n’ is the total numbers in the series.

The series of first ‘n’ even numbers is 2 + 4 + 6 + ………. + 2n

Here a = 2, d = 2  

$\text { Therefore Sum }\left(S_{e}\right)=\frac{n}{2}[2 \times 2+(n-1) \times 2]$

$\begin{array}{l}{=\frac{n}{2}[4+2 n-2]} \\\\ {=\frac{n}{2}[2+2 n]} \\\\ {=\frac{2 n+2 n^{2}}{2}} \\\\ {=\frac{2 n(1+n)}{2}=n(1+n)}\end{array}$

Thus sum of first n even numbers is n(1+n)

The series of first n odd numbers is 1 + 3 + 5 + …….+ n

Here a = 1, d = 2

$\begin{array}{l}{\text { Therefore Sum }\left(S_{o}\right)=\frac{n}{2}[2 \times 1+(n-1) \times 2]} \\\\ {=\frac{n}{2}[2+(n-1) \times 2]} \\\\ {=\frac{n}{2}[2+2 n-2]} \\\\ {=\frac{n}{2}[2 n]=n^{2}}\end{array}$

Thus the ratio of sum of first n even numbers and odd numbers is $\frac{n(n+1)}{n^{2}}=\frac{(n+1)}{n}$