Question

The number of terms in the expansion of ${(x^2+1+\frac{1}{x^2})}^n,n\in N$ is-

• A. $2n$
• B. $3n$
• C. $2n+1$
• D. $3n+1$

Answer 1

Answer: (C)

$2n+1$

We have the given expansions

$\left(x^2+1+\frac{1}{x^2}\right)$

$=1+x^2+\frac{1}{x^2}$

$=1+x^2+\frac{1}{x^2}+1-1$

$=2+x^2+\frac{1}{x^2}-1$

$={\left[{\left(x^2+\frac{1}{x^2}\right)}^2-1\right]}^n$

Now,

${}^n{C}_0{\left(x+\frac{1}{x}\right)}^n{+}^n{C}_1{\left(x+\frac{1}{x}\right)}^{n-1}{\left(-1\right)}^1+......{+}^n{C}_0{\left(x+\frac{1}{x}\right)}^n{+}^n{C}_n{\left(-1\right)}^n$

On solving the expression we get two constant term upto $n$

Hence

No. of terms $=n+n+1$

$=2n+1$