The greatest term in the expansion of sqrt 3 | vedclass

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

Question

(a) Let $(r + 1)^{th}$ term be the greatest term.

Then ${T_{r + 1}} = \sqrt 3 .{\,^{20}}{C_r}{\left( {\frac{1}{{\sqrt 3 }}} \right)^r}$and ${T_r} =

Vedclass · Accepted Answer

$\frac{{25840}}{9}$

Vedclass · Answer

(a) Let $(r + 1)^{th}$ term be the greatest term.

Then ${T_{r + 1}} = \sqrt 3 .{\,^{20}}{C_r}{\left( {\frac{1}{{\sqrt 3 }}} ight)^r}$and ${T_r} = \sqrt 3 .\,{\,^{20}}{C_{r - 1}}{\left( {\frac{1}{{\sqrt 3 }}} ight)^{r - 1}}$ 

Now $\frac{{{T_{r + 1}}}}{{{T_r}}} = \frac{{20 - r + 1}}{r}\left( {\frac{1}{{\sqrt 3 }}} ight)$ 

$	herefore \,\,\,\,{T_{r + 1}} \ge {T_r} \Rightarrow 20 - r + 1 \ge \sqrt 3 r$ 

$ \Rightarrow 21 \ge r(\sqrt 3 + 1)\,\, \Rightarrow r \le \frac{{21}}{{\sqrt 3 + 1}}$

$ \Rightarrow \,\,r \le 7.686 \Rightarrow r = 7$ 

Hence the greatest term is 

${T_8} = \sqrt 3 {\,^{20}}{C_7}{\left( {\frac{1}{{\sqrt 3 }}} ight)^7} = \frac{{25840}}{9}$

The greatest term in the expansion of $\sqrt 3 {\left( {1 + \frac{1}{{\sqrt 3 }}} \right)^{20}}$ is

Similar Questions

In the expansion of $(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to

The middle term in the expansion of ${\left( {1 - \frac{1}{x}} \right)^n}\left( {1 - {x}} \right)^n$ in powers of $x$ is

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}\,,\,\left( {x > 0} \right),$ are $m$ and $n$ respectively, then $\frac{m}{n}$ is equal to

Find the coefficient of $a^{4}$ in the product $(1+2 a)^{4}(2-a)^{5}$ using binomial theorem.

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is