Question

The number of terms containing $x^5$ in the expansion of $(2-x+3x^2{)}^6$, is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

${(3x^2-x+2)}^6={(3x^2+(2-x\left)\right)}^6$

$={}^6{C}_0{\left(3x^2\right)}^0{(2-x)}^6{+}^6{C}_1{\left(3x^2\right)}^1{(2-x)}^5{+}^6{C}_2{\left(3x^2\right)}^2{(2-x)}^4{+}^6{C}_3{\left(3x^2\right)}^3{(2-x)}^3+...{}^{}{.}^6{C}_6{\left(3x^2\right)}^6{(2-x)}^0$

$\sum \frac{6!}{a!b!c!}{\left({3x}^2\right)}^a{2}^b{(-x)}^c$

$a=2,c=1\Rightarrow x^5$

$a=1,c=3\Rightarrow x^5$

$a=0,c=5\Rightarrow x^5$

Therefore there are $3$ terms.