Question

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

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

Answer 1

The correct option is C $2n+1$

${(x^2+2+\frac{1}{x^2})}^n={(x^2+2.x.\frac{1}{x}+\frac{1}{x^2})}^n={\left({(x+\frac{1}{x})}^2\right)}^n={(x+\frac{1}{x})}^{2n}{=}_0^{2n}C.x^{2n}.{\left(\frac{1}{x}\right)}^0{+}_1^{2n}C.x^{2n-1}.{\left(\frac{1}{x}\right)}^1+.....{+}_{2n}^{2n}C.x^0.{\left(\frac{1}{x}\right)}^{2n}$

As there is positive sign between each term, hence no term will get cancel out.

Hence number of terms$=2n+1$