Question

If sum of the first n even natural numbers is S1 and sum of the first n odd
natural numbers is S2, then prove that n. S1 = (n+1)S2.​

Answer 1

Answer:

Sum  of the first even natural numbers:

• S1= 2+4+6+...+2n
• S1= n/2(2+2n)= n(n+1)

Sum of the first odd natural numbers:

• S2= 1+3+5+...+2n-1
• S2= n/2(1+2n-1)= n²

n*S1= n*n(n+1)= n²(n+1)

(n+1)*S2= (n+1)*n²= n²(n+1)

So n*S1=(n+1)*S2 as right sides of above 2 equations are same, hence proved