Question

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

• A. $40$
• B. $50$
• C. $30$
• D. $60$

Answer 1

The correct option is D $60$

${(1+x^2)}^5{(1+x)}^5=({{}^{5}C}_0{\left(x^2\right)}^{}+{{}^{5}C}_1{\left(x^2\right)}^{}+....+{{}^{5}C}_5{\left(x^2\right)}^{})(1+{{}^{4}C}_{1}x+{{}^{4}C}_2{\left(x^2\right)}^{}+....+{{}^{4}C}_4{\left(x^4\right)}^{})$

$(1+4x+6x^2+4x^3+x^4)$

$={\sum }_{r=0}^5{{}^{5}C}_r{\left(x^2\right)}^r={\sum }_{r=0}^5{{}^{5}C}_r{\left(x^2\right)}^r+4{\sum }_{r=0}^5{{}^{5}C}_r{\left(x^{2r+1}\right)}^{}+6{\sum }_{r=0}^5{{}^{5}C}_r{\left(x^{2r+2}\right)}^{}+4{\sum }_{r=0}^5{{}^{5}C}_r{\left(x^{2r+3}\right)}^{}+{\sum }_{r=0}^5{{}^{5}C}_r{\left(x^{2r+4}\right)}^{}$

so, coefficient of $x^5=0+4.{{}^{5}C}_2+6.0+4.{{}^{5}C}_1+0=4\times 10+4\times 5=40+20=60$