Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)^99. Let a be the middle term in the expansion of (2+ 1 √2)^200.…

In the binomial expansion of

${(1+x)}^{99}$(1+x)99, the sum of odd coefficients of $x$x is given by

$$K={C}_{1}+{C}_{3}+\ldots .+{C}_{99}=\frac{{2}^{99}-0}{2}={2}^{98}$$K=C1​+C3​+….+C99​=2299−0​=298

Also, $a=$a=Middle term in the binomial expansion of ${(2+\frac{1}{\sqrt{2}})}^{200}$(2+2​1​)200 is

$${T}_{\frac{200}{2}+1}={T}_{100+1}=C100200{(2)}^{100}{(\frac{1}{\sqrt{2}})}^{100}$$T2200​+1​=T100+1​=C100200(2)100(2​1​)100

$$\Rightarrow {T}_{100+1}=C100200\cdot {2}^{50}$$⇒T100+1​=C100200⋅250

So,

$$\frac{C99200K}{a}$$aC99200K​

$$=\frac{C99200\times {2}^{98}}{C100200\times {2}^{50}}=\frac{100}{101}\times {2}^{48}$$=C100200×250C99200×298​=101100​×248

$$=\frac{25}{101}\times {2}^{50}=(\frac{m}{n}){2}^{l}$$=10125​×250=(nm​)2l

On comparing, we get

$$l=50,m=25\mathrm{and}n=101$$l=50,m=25andn=101

Therefore, $(l,n)\equiv (50,101)$(l,n)≡(50,101).