What is the constant term in the binomial exp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given expression: $(1+3x)^n \left(1+\frac{1}{3x}\right)^n$ $= (1+3x)^n \left(\frac{3x+1}{3x}\right)^n$ $= \frac{(1+3x)^n (1+3x)^n}{(3x)^n}$ $= \fr

Vedclass · Accepted Answer

$\binom{2n}{n}$

Vedclass · Answer

(A) Given expression: $(1+3x)^n \left(1+\frac{1}{3x}ight)^n$ $= (1+3x)^n \left(\frac{3x+1}{3x}ight)^n$ $= \frac{(1+3x)^n (1+3x)^n}{(3x)^n}$ $= \frac{(1+3x)^{2n}}{(3x)^n}$ The general term in the expansion of $(1+3x)^{2n}$ is $T_{r+1} = \binom{2n}{r} (3x)^r$. To find the constant term,we need the term where the power of $x$ is $0$. The expression is $\frac{1}{(3x)^n} 	imes \sum_{r=0}^{2n} \binom{2n}{r} (3x)^r = \sum_{r=0}^{2n} \binom{2n}{r} \frac{(3x)^r}{(3x)^n} = \sum_{r=0}^{2n} \binom{2n}{r} (3x)^{r-n}$. The constant term occurs when $r-n = 0$,i.e.,$r = n$. Substituting $r=n$,the constant term is $\binom{2n}{n} (3x)^{n-n} = \binom{2n}{n}$. Thus,the constant term is $\binom{2n}{n}$.

What is the constant term in the binomial expansion of $(1+3x)^n \left(1+\frac{1}{3x}\right)^n$?

Similar Questions

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is .... .

The positive value of $a$ such that the coefficient of $x^5$ is equal to that of $x^{15}$ in the expansion of $(x^2 + \frac{a}{x^3})^{10}$ is

If the maximum value of the term independent of $t$ in the expansion of $\left( t^{2} x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}$,$x \geq 0$,is $K$,then $8K$ is equal to $....$

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0$, if the term independent of $x$ is $(221)k$, then $k$ is equal to:

In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module