Question

Coefficient of $x^5$ in the expansion of ${(1+x^2)}^5{(1+x)}^4$ is:

• A. $61$
• B. $59$
• C. $0$
• D. $60$

Answer 1

The correct option is D $60$

Co-efficient of $x^5$ in $(1+x^2{)}^5(1+x{)}^4$

Binomial expansion of

$(a+b{)}^n{=}^n{C}_a{a}^n{+}^n{C}_1{a}^{n-1}{b}^1+..{.}^n{C}_2{a}^{n-r}{b}^r+...$

$\therefore (1+x^2{)}^5(1+x{)}^4$

${(}^5{C}_0\left(1\right){+}^5{C}_1\left(1\right)x^2{+}^5{C}_2\left(1\right)\left(x^2{)}^2{+}^5{C}_3\right(x^2{)}^3{+}^5{C}_4\left(x^2{)}^4{+}^5{C}_c\right(x^2{)}^5)$

$\times {(}^4{C}_0\left(1\right){+}^4{C}_1\left(x\right){+}^4{C}_2\left(x{)}^2{+}^4{C}_3\right(x{)}^3{+}^4{C}_4\left(x{)}^4\right)$

$\therefore \left\{{}^5{C}_2{x}^4{\times }^4{C}_{1}x{+}^5{C}_{1}x{+}^5{C}_1\right(x^2\left){\times }^4{C}_3\right(x{)}^3\}$

$\therefore$ coefficient of $x^5$

$\Rightarrow (\frac{5\times 4}{2\times 1}\times 4)x^5+(5\times 4)x^5$

$\Rightarrow (10\times 4+20)x^5$

$\Rightarrow 60x^5$

$\therefore$ coefficient of $x^5\Rightarrow 60$ (option D)