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(D) The binomial expansion of $(1+x)^n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + \frac{n(n-1)\dots(n-r+1)}{r!}x^r + \dots$ Here,$n = \frac

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$8$

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(D) The binomial expansion of $(1+x)^n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + \frac{n(n-1)\dots(n-r+1)}{r!}x^r + \dots$ Here,$n = \frac{27}{5} = 5.4$. Since $x > 0$,the terms will be negative if the coefficient of $x^r$ is negative. The general term is $t_{r+1} = \binom{n}{r} x^r = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$. We check the signs of the coefficients: For $r=1$: $\binom{5.4}{1} = 5.4 > 0$. For $r=2$: $\binom{5.4}{2} = \frac{5.4 	imes 4.4}{2} = 11.88 > 0$. For $r=3$: $\binom{5.4}{3} = \frac{5.4 	imes 4.4 	imes 3.4}{3 	imes 2 	imes 1} = 13.464 > 0$. For $r=4$: $\binom{5.4}{4} = \frac{5.4 	imes 4.4 	imes 3.4 	imes 2.4}{4 	imes 3 	imes 2 	imes 1} = 8.0784 > 0$. For $r=5$: $\binom{5.4}{5} = \frac{5.4 	imes 4.4 	imes 3.4 	imes 2.4 	imes 1.4}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 2.261952 > 0$. For $r=6$: $\binom{5.4}{6} = \frac{5.4 	imes 4.4 	imes 3.4 	imes 2.4 	imes 1.4 	imes 0.4}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 0.1507968 > 0$. For $r=7$: $\binom{5.4}{7} = \frac{5.4 	imes 4.4 	imes 3.4 	imes 2.4 	imes 1.4 	imes 0.4 	imes (-0.6)}{7!} Since the term $t_{r+1}$ is negative for $r=7$,the first negative term is $t_{7+1} = t_8$. Thus,$k=8$.

If $x$ is a positive real number and the first negative term in the expansion of $(1+x)^{27/5}$ is $t_k$,then $k=$

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