Question

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

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

Answer 1

Answer: (C)

$2n+1$

$(x^2+1+\frac{1}{x^2}{)}^n$
$=(x^2+2+\frac{1}{x^2}-1{)}^n$
$=\left[\right(x+\frac{1}{x}{)}^2-1{]}^n$
$=\left[\right(x+\frac{1}{x}+1\left)\right(x+\frac{1}{x}-1){]}^n$
$=(x+\frac{1}{x}+1{)}^n((x+\frac{1}{x}-1{)}^n$
Hence there will be 2n+1 terms.
Alternatively, considering $(x^2+1+\frac{1}{x^2}{)}^n$
If we replace the 1 by 2, we will get $2n$ terms dependent on x, since one term will be independent of $x$. However, since $x^2+1+\frac{1}{x^2}$ is not a perfect square, there won't be any term independent of $x$ except 1 itself. Hence there will $2n+1$ terms.