a ની કઈ કિમત માટે {left( {{x^2},, + ,,fr | vedclass

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

Question

$\left(x^{2}+\frac{a}{x^{3}}\right)^{10}$

As we know that the general term $\left(T_{r+1}\right)$ of the expression $(a+b)^{n}$ is-

$T_{r+1}={ }

Vedclass · Accepted Answer

$\frac{1}{{2\,\sqrt 3 }}$

Vedclass · Answer

$\left(x^{2}+\frac{a}{x^{3}}ight)^{10}$

As we know that the general term $\left(T_{r+1}ight)$ of the expression $(a+b)^{n}$ is-

$T_{r+1}={ }_{n} C_{r} a^{n-r} b^{n}$

Therefore, in the given expansion $\left(x^{2}+\frac{a}{x^{3}}ight)^{10}$ $a=x^{2}$

$b=\frac{a}{x^{3}}$

$n=10$

$	herefore$ general term of the given expression-

$T_{r+1}={ }_{10} C_{r} x^{20-2 r} \frac{a^{r}}{x^{3 r}}$

$\Rightarrow T_{r+1}={ }_{10} C_{r} x^{20-5 r} a^{r}$

For coefficient of $x^{5}-$

$20-5 r=5$

$5 r=15 \Rightarrow r=3$

$	herefore T_{4}={ }_{10} C_{3} x^{5} a^{3}$

For coefficient of $x^{15}-$

$20-5 r=15$

$5 r=5 \Rightarrow r=1$

$	herefore T_{2}={ }_{10} C_{1} x^{15} a$

Given that the coefficient of $x^{5}$ is equal to the coefficient of $x^{15}$, i.e.,

${ }_{10} C_{3} a^{3}={ }_{10} C_{1} a \ldots$

As we know that,

${ }_{n} C_{r}=\frac{n !}{r !(n-r) !}$

$	herefore{ }_{10} C_{3}=\frac{10 !}{3 !(10-3) !}=120$

${ }_{10} C_{1}=\frac{10 !}{1 !(10-1) !}=10$

Now from $e q^{n}(1)$, we have

$120 	imes a^{3}=10 a$

$\Rightarrow a^{2}=\frac{1}{12}$

$\Rightarrow a=\pm \frac{1}{2 \sqrt{3}}$

As we have to find positive value of $a$. $	herefore a=\frac{1}{2 \sqrt{3}}$

$a$ ની કઈ કિમત માટે ${\left( {{x^2}\,\, + \,\,\frac{a}{{{x^3}}}} \right)^{10}}$ ના વિસ્તરણમાં $x^5$ અને $x^{15}$ નો સહગુણકો સમાન થાય ?

Similar Questions

$\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}$ ના વિસ્તરણનું $x$ થી સ્વતંત્ર પદ(અચળ પદ) શોધો.

$\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}$ ના વિસ્તરણમાં અચળપદ મેળવો.

$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$ નાં વિસ્તરણમાં અચળ પદ જો $\alpha \times 2^8 \times \sqrt[5]{3}$ હોય, તો $25 \alpha=$...............

${\left( {2x + \frac{1}{{3x}}} \right)^6}$ ના વિસ્તરણમાં અચળપદ મેળવો.

${\left( {x - \frac{3}{{{x^2}}}} \right)^9}$ ના વિસ્તરણમાં અચળપદ મેળવો.