The coefficient of x^{n-6} in the expansion n | vedclass

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Question

(A) The general term in the product is of the form $\left[ \frac{x}{k} - \frac{^nC_{k-1} + ^nC_k}{^nC_{k-1}} \right]$.Using the identity $\frac{^nC_k}

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$^nC_6 (n+1)^6$

Vedclass · Answer

(A) The general term in the product is of the form $\left[ \frac{x}{k} - \frac{^nC_{k-1} + ^nC_k}{^nC_{k-1}} \right]$.Using the identity $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$,we have $\frac{^nC_{k-1} + ^nC_k}{^nC_{k-1}} = 1 + \frac{n-k+1}{k} = \frac{k+n-k+1}{k} = \frac{n+1}{k}$.Thus,the expression becomes $n! \left( x - (n+1) \right) \left( \frac{x}{2} - \frac{n+1}{2} \right) \dots \left( \frac{x}{n} - \frac{n+1}{n} \right)$.Factoring out $\frac{1}{k}$ from each bracket,we get $n! \cdot \frac{1}{1 \cdot 2 \cdot 3 \dots n} (x - (n+1))^n$.Since $n! = 1 \cdot 2 \cdot 3 \dots n$,the expression simplifies to $(x - (n+1))^n$.The coefficient of $x^{n-6}$ in $(x - (n+1))^n$ is given by the binomial theorem as $^nC_6 (x)^{n-6} (-(n+1))^6 = ^nC_6 (n+1)^6 x^{n-6}$.Therefore,the coefficient is $^nC_6 (n+1)^6$.

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