Question

Find the sum of first 100 odd natural numbers of an ap

Answer 1

hey
here is ur answer ......
We know that sum of cubes of first n natural numbers is = (n(n+1)/2)2.
Sum of cube of first n even natural numbers 
23 + 43 + 63 + ……… + (2n)3
Even Sum = 23 + 43 + 63 + .... + (2n)3 if we multiply by 23 then = 23 x (13 + 23 + 32 + .... + (n)3) = 23 + 43 + 63 + ......... + (2n)3 = 23 (n(n+1)/2)2 = 8(n(n+1))2/4 = 2(n(n+1))2

Answer 2

1,3,5..

a=1

d=2

n=100

Sn=n/2[2a+(n-1)d]

S100=100/2[2*1+(100-1)2]

=50[2+99*2]

=50[2+198]

=50*200

=10,000