If x is positive,the first negative term in t | vedclass

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(C) The general term $T_{r+1}$ in the expansion of $(1+x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$.For the term to be negativ

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$8^{th}$ term

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(C) The general term $T_{r+1}$ in the expansion of $(1+x)^n$ is given by $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$.For the term to be negative,the product $n(n-1)(n-2)\dots(n-r+1)$ must be negative since $x^r$ and $r!$ are positive for $x > 0$.Here $n = \frac{27}{5} = 5.4$.The terms become negative when the factor $(n-r+1) $5.4 - r + 1 Since $r$ must be an integer,the smallest integer $r$ is $7$.Thus,the first negative term is $T_{7+1} = T_8$,which is the $8^{th}$ term.

If $x$ is positive,the first negative term in the expansion of $(1 + x)^{27/5}$ is

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