If the coefficients of x−2 and x−4 in the expansion of ( x^1 \over 3 + 1 \over 2x^1 \over 3 )^18,( x \gt 0 ), are m and n respectively, then m \over n is equal to :

Binomial Theorem PYQ — JEE Main 2016 · 10 Apr · Shift 1 · Q23 with Solution

View written solutionFree

Correct answer: A

Write the general term

For $\left(x^{1/3}+\frac{1}{2x^{1/3}}\right)^{18},$ the general term is $T_{r+1}=\binom{18}{r}\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^r.$

Simplifying,

\binom{18}{r}\frac{1}{2^r}x^{\frac{18-2r}{3}}.$$ So the exponent of $x$ in the general term is $$\frac{18-2r}{3}.$$ --- 2. **Find coefficient of $x^{-2}$** We need $$\frac{18-2r}{3}=-2.$$ So, $$18-2r=-6$$ $$2r=24$$ $$r=12.$$ Hence the coefficient of $x^{-2}$ is $$m=\binom{18}{12}\frac{1}{2^{12}}=\binom{18}{6}\frac{1}{2^{12}}.$$ --- 3. **Find coefficient of $x^{-4}$** We need $$\frac{18-2r}{3}=-4.$$ So, $$18-2r=-12$$ $$2r=30$$ $$r=15.$$ Hence the coefficient of $x^{-4}$ is $$n=\binom{18}{15}\frac{1}{2^{15}}=\binom{18}{3}\frac{1}{2^{15}}.$$ --- 4. **Compute $\dfrac{m}{n}$** $$\frac{m}{n}=\frac{\binom{18}{6}/2^{12}}{\binom{18}{3}/2^{15}}=2^3\cdot \frac{\binom{18}{6}}{\binom{18}{3}}.$$ Now, $$\binom{18}{6}=\frac{18!}{6!12!}, \qquad \binom{18}{3}=\frac{18!}{3!15!}.$$ Thus, $$\frac{\binom{18}{6}}{\binom{18}{3}}=\frac{3!15!}{6!12!}= rac{15\cdot14\cdot13}{6\cdot5\cdot4}.$$ So, $$\frac{m}{n}=8\cdot \frac{15\cdot14\cdot13}{6\cdot5\cdot4}.$$ Simplify: $$\frac{15\cdot14\cdot13}{6\cdot5\cdot4}= rac{2730}{120}=\frac{91}{4}.$$ Therefore, $$\frac{m}{n}=8\cdot \frac{91}{4}=2\cdot91=182.$$ --- 5. **Check options** The correct value is $$\frac{m}{n}=182.$$ So the correct option is **A**.

More from Binomial Theorem

The term independent of $x$ in expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ is2013 · MCQ
If $n$ is a positive integer, then $(3 + 1)^{2 n} - (3 - 1)^{2 n}$ is :2012 · MCQ
The remainder left out when $8^{2 n} - (62)^{2 n + 1}$ is divided by 9 is :2009 · MCQ
In the binomial expansion of $(a - b)^{n}, n \geq 5,$ the sum of $5^{t h}$ and $6^{t h}$ terms is zero, then $a / b$ equals2007 · MCQ
If the expansion in powers of $x$ of the function $\frac{1}{( 1 - a x ) ( 1 - b x )}$ is $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} .....$ then $a_{n}$ is2006 · MCQ
If $x$ is so small that $x^{3}$ and higher powers of $x$ may be neglected, then $\frac{( 1 + x ) ^{\frac{3}{2}} - ( 1 + \frac{1}{2} x ) ^{3}}{( 1 - x ) ^{\frac{1}{2}}}$ may be approximated…2005 · MCQ
If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then m and r satisfy the equation2005 · MCQ
If the coefficient of $x^{7}$ in $[a x^{2} + (\frac{1}{b x})]^{11}$ equals the coefficient of $x^{- 7}$ in $[a x - (\frac{1}{b x ^{2}})]^{11}$ , then $a$ and $b$ satisfy…2005 · MCQ

View written solutionFree

Correct answer: A

Write the general term

For $\left(x^{1/3}+\frac{1}{2x^{1/3}}\right)^{18},$ the general term is $T_{r+1}=\binom{18}{r}\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^r.$

Simplifying,

\binom{18}{r}\frac{1}{2^r}x^{\frac{18-2r}{3}}.$$ So the exponent of $x$ in the general term is $$\frac{18-2r}{3}.$$ --- 2. **Find coefficient of $x^{-2}$** We need $$\frac{18-2r}{3}=-2.$$ So, $$18-2r=-6$$ $$2r=24$$ $$r=12.$$ Hence the coefficient of $x^{-2}$ is $$m=\binom{18}{12}\frac{1}{2^{12}}=\binom{18}{6}\frac{1}{2^{12}}.$$ --- 3. **Find coefficient of $x^{-4}$** We need $$\frac{18-2r}{3}=-4.$$ So, $$18-2r=-12$$ $$2r=30$$ $$r=15.$$ Hence the coefficient of $x^{-4}$ is $$n=\binom{18}{15}\frac{1}{2^{15}}=\binom{18}{3}\frac{1}{2^{15}}.$$ --- 4. **Compute $\dfrac{m}{n}$** $$\frac{m}{n}=\frac{\binom{18}{6}/2^{12}}{\binom{18}{3}/2^{15}}=2^3\cdot \frac{\binom{18}{6}}{\binom{18}{3}}.$$ Now, $$\binom{18}{6}=\frac{18!}{6!12!}, \qquad \binom{18}{3}=\frac{18!}{3!15!}.$$ Thus, $$\frac{\binom{18}{6}}{\binom{18}{3}}=\frac{3!15!}{6!12!}= rac{15\cdot14\cdot13}{6\cdot5\cdot4}.$$ So, $$\frac{m}{n}=8\cdot \frac{15\cdot14\cdot13}{6\cdot5\cdot4}.$$ Simplify: $$\frac{15\cdot14\cdot13}{6\cdot5\cdot4}= rac{2730}{120}=\frac{91}{4}.$$ Therefore, $$\frac{m}{n}=8\cdot \frac{91}{4}=2\cdot91=182.$$ --- 5. **Check options** The correct value is $$\frac{m}{n}=182.$$ So the correct option is **A**.

Binomial Theorem question

Binomial Theorem question

More from Binomial Theorem

Binomial Theorem question

Binomial Theorem question

More from Binomial Theorem