(In reply to
Answer by K Sengupta)
We know that:
S(n) = (n/2)*{2a+(n-1)*d}, where:
S(n) = Sum of the given series to n terms
a = First term
d= common difference
In the given problem, we have:
a= 1 and, d=2, so that:
S(n)
= (n/2)*{2*1+(n-1)*2}}
= (n/2)*{2+2n-2}
= (n/2)*(2n)
= n^2
Consequently, the sum of the consecutive odd numbers beginning at 1 always adds up to perfect square.