Question

Show that the sum of first n even natural numbers is equal to 1 + 1 by n times the sum of first n odd natural numbers

Answer 1

Answer:

Step-by-step explanation:

Sum of n natural numbers = 1/2. (1 + 1/n). Here is the proof.

Sum of ’n’ natural numbers = 1/2.n(n+1)

Sum of ’n’ natural odd numbers = n^2

1/2.n(n+1) = n^2 x (m), where ‘m’ is the multiplier.

m = 1/2. n(n+1) ÷ n^2

m = 1/2(1 + 1/n)

Verification.

Sum of 3 natural numbers = 1 + 2 + 3 = 6

Sum of 3 natural odd numbers = 1 + 3 + 5 = 9

1/2(1 + 1/n) x 9 = 1/2 x (1 + 1/3) x 9 = 1/2 x 4/3 x 9 = 2 x 3 = 6 (Verified)

Answer 2

Answer:

n+n^2=n^2+n

Step-by-step explanation:

First n even natural number are

2,4,6,8....n. a=2 d=2

First n odd natural number are

1,3,5,7,9...n a=1 d=2

Let Sum of first n even number be

Se=n/2[2×2+(n+1)2]

Let Sum of first n odd even number be So=n/2[2+(n+1)2]

To prove :

Se=[1+1/n]So

Se=[n+1/n]So

n/2[4+2n-2]={n+1/n[n/2(2+2n-2)]}

n/2(2+2n)=[n+1/n(n/2×2n])

n+n^2=n+1/n[n^2]

n+n^2=n^2+n

Hence proved