Question

The term independent of $x$, in the expansion of ${(1+\frac{1}{x}+x+x^2)}^4$ is

• A. $35$
• B. $30$
• C. $32$
• D. $31$

Answer 1

The correct option is D $31$

${[1+\frac{1}{x}+x+x^2]}^4$
$=\frac{{[1+x+x^2+x^3]}^4}{x^4}$
$=\frac{{[(1+x)+x^2(x+1)]}^4}{x^4}$
$\frac{{(1+x)}^4{(1+x^2)}^4}{x^4}⟶\left(1\right)$
Term independent $x$ in ${[1+\frac{1}{x}+x+x^2]}^4$
=Co efficient of $x^4$ in equation 1 [from equation 1]
${(1+x)}^4{(1+x^2)}^4=\left[{}^4{C}_0{+}^4{C}_{1}x{+}^4{C}_2{x}^2{+}^4{C}_3{x}^3{+}^4{C}_4{x}^4\right].\left[{}^4{C}_0{+}^4{C}_1{x}^2{+}^4{C}_2{x}^4{+}^4{C}_3{x}^6{+}^4{C}_4{x}^8\right]$
Co efficient of $x^4{=}^4{C}_0{\times }^4{C}_2{+}^4{C}_2{\times }^4{C}_1{+}^4{C}_4{\times }^4{C}_0$
$=1\times 6+6\times 4+1\times 1$
$=6+24+1$
$=31$
$\therefore$ Term independent of $x=31$