If the coefficents of {x^3} and {x^4} in | vedclass

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

Question

In the expansion of $\left(1+a x+b x^{2}\right)(1-2 x)^{18},$

Coefficient of $x^{3}$ in $\left(1+a x+b x^{2}\right)(1-2 x)^{18}$

$=$ Coefficient

Vedclass · Accepted Answer

($16$,$\frac{{272}}{3}$)

Vedclass · Answer

In the expansion of $\left(1+a x+b x^{2}ight)(1-2 x)^{18},$

Coefficient of $x^{3}$ in $\left(1+a x+b x^{2}ight)(1-2 x)^{18}$

$=$ Coefficient of $x^{3}$ in $(1-2 x)^{18}$

${+	ext { Coefficient of } x^{2} 	ext { in a }(1-2 x)^{	ext {18 }}}$

${	ext { + Coefficient of } x 	ext { in } b(1-2 x)^{	ext {18 }}}$

$=-^{18} \mathrm{C}_{3} \cdot 2^{3}+\mathrm{a}^{18} \mathrm{C}_{2} \cdot 2^{2}-\mathrm{b}^{18} \mathrm{C}_{1} \cdot 2$

$\because-^{18} C_{3} \cdot 2^{3}+a^{18} C_{2} \cdot 2^{2}-b^{18} C_{1} \cdot 2=0$

$\Rightarrow \frac{18 	imes 17 	imes 16}{3 	imes 2} \cdot 8+a \cdot \frac{18 	imes 17}{2} 2^{2}-b \cdot 18 \cdot 2=0$

$\Rightarrow 17 a-b=\frac{34 	imes 16}{3}..........(i)$

Similarly, coefficient of $x^{4}$

$^{18} \mathrm{C}_{4} \cdot 2^{4}-\mathrm{a} \cdot^{18} \mathrm{C}_{3} 2^{3}+b \cdot^{18} \mathrm{C}_{2} \cdot 2^{2}=0$

$32 a-32 b=240..........(ii)$

On solving Eqs. $(i)$ and $(ii)$, we get

$a=16 	ext { and } b=\frac{272}{3}$

If the coefficents of ${x^3}$ and ${x^4}$ in the expansion of $\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$ in powers of $x$ are both zero, then $ (a,b) $ is equal to

Similar Questions

If the third term in the binomial expansion of ${\left( {1 + {x^{{{\log }_2}\,x}}} \right)^5}$ equals $2560$, then a possible value of $x$ is

Find the term independent of $x$ in the expansion of $\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x\,>\,0$

If $7^{th}$ term from beginning in the binomial expansion ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln \,x} \right)^9},\,x > 0$ is equal to $729$ , then possible value of $x$ is

Find the $13^{\text {th }}$ term in the expansion of $\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0$

The middle term in the expression of ${\left( {x - \frac{1}{x}} \right)^{18}}$ is