Question

The sum of co-effcients in the expansion of ${(\frac{1}{x}+2x)}^n$ is equal to $729$. Then, the constant term in the expansion is :

• A. 150
• B. 160
• C. 170
• D. 180

Answer 1

Answer: (B)

160

$(1+x{)}^n={}^n{C}_0^n{C}_{P}x+{}^n{C}_2{x}^2+......{.}^n{C}_n{x}^n$

part $x=2$

$(1+2{)}^n={}^n{C}_0+{}^n{C}_1(2)+2^2{}^n{C}_2+....{.}^{2^n}{C}_n$....(1)

$(\frac{1}{x}+2x{)}^n={\left(\frac{1}{x}\right)}^n{}^n{C}_o{+}^n{C}_1{\left(\frac{1}{x}\right)}^{n-1}(2x)+....{.}^n{C}_n(2x{)}^n$

Sum of coefficient $={}^n{C}_o+2{}^n{C}_1+.....{.2}^n{}^n{C}_x$

$729=(1+2{)}^n$ from (1)

$729=3^n$

$\Rightarrow 3^6=3^n\Rightarrow n=6$

${(\frac{1}{x}+2x)}^6={\left(\frac{1}{x}\right)}^6{}^6{C}_0+{}^6{C}_1{\left(\frac{1}{x}\right)}^5(2x{)}^1+{}^6{C}_2{\left(\frac{1}{x}\right)}^4(2x{)}^2+$

${}^6{C}_3{\left(\frac{1}{x}\right)}^3(2x{)}^3+{}^6{C}_4{\left(\frac{1}{x}\right)}^2(2x{)}^4+{}^6{C}_4{\left(\frac{1}{x}\right)}^2(2x{)}^5+{}^6{C}_6(2x{)}^6$

Constant term $={}^6{C}_3\times 2^3$

$=\frac{6!}{3!3!}\times 2\times 2\times 2$

$\Rightarrow \frac{720}{6\times 6}\times 2\times 2\times 2$

$\Rightarrow 160$

The constant term value is $160$