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)

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