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Binomial Theorem question

2022 · 29 Jun · Shift 2 · Q40
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  5. /2022 · 29 Jun · Shift 2 · Q40

Binomial Theorem question

2022 · 29 Jun · Shift 2 · Q40

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
Let the coefficients of x −-− 1 and x −-− 3 in the expansion of (2x15−1x15)15,x>0{\left( {2{x^{{1 \over 5}}} - {1 \over {{x^{{1 \over 5}}}}}} \right)^{15}},x \gt 0(2x51​−x51​1​)15,x>0, be m and n respectively. If r is a positive integer such that mn2=15Cr . 2rm{n^2} = {}^{15}{C_r}\,.\,{2^r}mn2=15Cr​.2r, then the value of r is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Write the general term of (2x1/5−x−1/5)15.\left(2x^{1/5}-x^{-1/5}\right)^{15}.(2x1/5−x−1/5)15.

Using the binomial theorem, Tk+1=(15k)(2x1/5)15−k(−x−1/5)k,k=0,1,2,…,15.T_{k+1}=\binom{15}{k}(2x^{1/5})^{15-k}(-x^{-1/5})^k,\qquad k=0,1,2,\dots,15.Tk+1​=(k15​)(2x1/5)15−k(−x−1/5)k,k=0,1,2,…,15.

So,

\binom{15}{k}2^{15-k}(-1)^k x^{\frac{15-2k}{5}}.$$ Thus the power of $x$ in the general term is $$\frac{15-2k}{5}=3-\frac{2k}{5}.$$ --- 2. **Find the coefficient of $x^{-1}$**. We need $$\frac{15-2k}{5}=-1.$$ So, $$15-2k=-5\implies 2k=20\implies k=10.$$ Hence, $$m=\binom{15}{10}2^{5}(-1)^{10}=\binom{15}{10}2^5.$$ --- 3. **Find the coefficient of $x^{-3}$**. We need $$\frac{15-2k}{5}=-3.$$ So, $$15-2k=-15\implies 2k=30\implies k=15.$$ Hence, $$n=\binom{15}{15}2^0(-1)^{15}=-1.$$ --- 4. **Compute $mn^2$**. Since $n=-1$, we have $$n^2=1.$$ Therefore, $$mn^2=m=\binom{15}{10}2^5.$$ Given that $$mn^2=\binom{15}{r}2^r,$$ we compare: $$\binom{15}{r}2^r=\binom{15}{10}2^5.$$ Now, $$\binom{15}{10}=\binom{15}{5}.$$ So, $$\binom{15}{r}2^r=\binom{15}{5}2^5.$$ Thus the matching positive integer is $$r=5.$$ --- 5. **Final answer** $$\boxed{5}$$
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