If the (r + 1)^{th} term is the first negativ | vedclass

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(A) The general term in the expansion of $(1 + x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)...(n-r+1)}{r!} x^r$.Here,$n = \frac{7}{2}$.The terms bec

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$5$

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(A) The general term in the expansion of $(1 + x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)...(n-r+1)}{r!} x^r$.Here,$n = \frac{7}{2}$.The terms become negative when the product of the factors in the numerator becomes negative.The factors are $\frac{7}{2}, \frac{5}{2}, \frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, ...$.The first negative factor occurs at the $5^{th}$ term,where the factor is $(n-r+1) = \frac{7}{2} - r + 1$.For the term to be negative,we require $\frac{7}{2} - r + 1  \frac{9}{2} = 4.5$.Since $r$ must be an integer,the smallest value is $r = 5$.

If the $(r + 1)^{th}$ term is the first negative term in the expansion of $(1 + x)^{7/2}$,then the value of $r$ is

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