Which is greater, D or N, and by how much?

D is the sum of the odd numbers from 1 through 99 inclusive, and N is the sum of the even numbers from 2 through 98 inclusive:
           D = 1+3+5+ ........+99 and N = 2+4+6+ ...... +98
Which is greater, D or N, and by how much?

MetroEye: Refer Progression03 for formulae.

Number of terms in odd number series = \(\dfrac{t_n+1}{2}\)= \(\dfrac{99+1}{2}\) = \(\dfrac{100}{2}\)  = 50

Number of terms in even number series = \(\dfrac{t_n}{2} = \dfrac{98}{2}\)  = 49

Sum, D = \( n^2\) = 50*50

Sum, N = n(n+1) = 49*50

=> D is greater than N.

Difference between D and N = D-N = 50*50-49*59 = 50(50-49) = 50

Hence D is greater than N by 50.