Question

The constant term in the expansion of ${(x^2+\frac{1}{x^2}+y+\frac{1}{y})}^8$ is :

• A. $4900$
• B. $4950$
• C. $5050$
• D. $5151$

Answer 1

Answer: (A)

$4900$

We have,

$\frac{8!}{\alpha !\beta !\gamma !\delta !}{\left(x^2\right)}^{\alpha }{\left(\frac{1}{x^2}\right)}^{\beta }{\left(y\right)}^{\gamma }{\left(\frac{1}{y}\right)}^{\delta }\frac{8!}{\alpha !\beta !\gamma !\delta !}{\left(x\right)}^{2\alpha -2\beta }\times {\left(y\right)}^{\gamma -\delta }$

$Forconstantterm\alpha =\beta ;\gamma =\delta \alpha +\beta +\gamma +\delta =82\alpha +2\gamma =8\alpha +\gamma =4\left(\beta =1\right)\alpha =1\gamma =3\left(\delta =3\right)\left(\beta =2\right)\alpha =2\gamma =2\left(\delta =2\right)\left(\beta =3\right)\alpha =3\gamma =1\left(\delta =1\right)\left(\beta =4\right)\alpha =4\gamma =0\left(\delta =0\right)\left(\beta =0\right)\alpha =0\gamma =4\left(\delta =4\right)$

$Cons\tan tterm=\frac{8!}{1!1!3!3!}+\frac{8!}{2!2!2!2!}+\frac{8!}{3!3!1!1!}+\frac{8!}{4!4!}=2\times \frac{8\times 7\times 6\times 5\times 4}{6}+\frac{8\times 7\times 6\times 120}{16}+\frac{8\times 7\times 6\times 5\times 4!}{24}\times 2=2240+2520+4\times 7\times 6=2240+2520+140=4900$

$Hence,theoptionAisthecorrectanswer.$