Question

The constant term in the expansion of $[1-(x-2)^2]^{10}$ is equal to

• A. $2^{10}$
• B. $6^{10}$
• C. $4^{10}$
• D. $5^{10}$
• E. $3^{10}$

Answer 1

Answer: (E)

$\left[1-(x-2)^{2}\right]^{10} =\left[1-\left(x^{2}-4 x+4\right)\right]^{10} $
$=\left[-3-\left(x^{2}-4 x\right)\right]^{10} $
$=(-1)^{10}\left[3+\left(x^{2}-4 x\right)\right]^{10} $
$=\left[3+\left(x^{2}-4 x\right)\right]^{10}$
$\therefore $ The general term is
$T_{r+1}={ }^{10} C_{r} 3^{10-r}\left(x^{2}-4 x\right)^{r}$
For constant term, put $r=0$
Hence, the constant term is $3^{10}$.