In the binomial expansion of
(1+x)99, the sum of odd coefficients of x is given by
K=C1+C3+….+C99=2299−0=298
Also, a=Middle term in the binomial expansion of (2+21)200 is
T2200+1=T100+1=C100200(2)100(21)100
⇒T100+1=C100200⋅250
So,
aC99200K
=C100200×250C99200×298=101100×248
=10125×250=(nm)2l
On comparing, we get
l=50,m=25andn=101
Therefore, (l,n)≡(50,101).