c for the diff values of r



r=1

a,b = 1,4/3

for r=2

c,d=2,8/3

for r=3

e,f= 3,12/3

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similarly the last pair wud be 120,480/3

so b+d+f....=4/3[1+2+3+...120]

a+c+e....=1+2+3+...120

so b+d+f.... - (a+c+e...)=

4/3(120)(121)/2 - 120(121)/2 [using sum of n terms]

=1/3(120)(121)/2

=2420