Question

14. Prove that the sum of the first n odd natural numbers and the sum of the first n even numbers is the ratio of n: n + 1. .

Answer 1

The first n odd natural numbers are 1, 3, 5, ... 2n-1.

first term 1

common difference 2

Last term (2n-1).

The sum is given by S = 1+3+5+...+(2n-1)

Using S = n (First term + Last term)/2

we get S = n (1+2n-1)/2

S = n² -----------------(1)

Now,

Sum of first n even natural numbers =

S = 2+ 4+...+ 2n = 2(1+2+3+....+n) = 2[n(n+1)/2] = n(n+1)

so, required ratio = n²:n(n+1) = n:(n+1) (Proved)

[ mark as brainlist if you understand ]

Answer 2

Step-by-step explanation:

Formula for :

Sum of first n odd natural numbers,

S1 = n * n

Sum of first n even natural numbers,

S2 = n * (n+1)

S1:S2 = n * n / n * (n+1)

One n gets cancelled, we get

S1:S2 = n : (n+1)