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(C) The binomial expansion of $(1+x)^n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ For $n = \frac{21}{5}$,the ter

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$\frac{-616}{5^7}$

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(C) The binomial expansion of $(1+x)^n$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ For $n = \frac{21}{5}$,the terms are $T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r$. The coefficients are positive as long as $(n-r+1) > 0$. We check the signs of the factors: $n = \frac{21}{5} = 4.2$ $n-1 = 3.2$ $n-2 = 2.2$ $n-3 = 1.2$ $n-4 = 0.2$ $n-5 = -0.8$ The first negative term occurs when $r=5$,which corresponds to the coefficient of $x^5$. The coefficient is $\frac{\frac{21}{5} 	imes \frac{16}{5} 	imes \frac{11}{5} 	imes \frac{6}{5} 	imes \frac{1}{5}}{5!} = \frac{21 	imes 16 	imes 11 	imes 6 	imes 1}{5^5 	imes 120} = \frac{22176}{3125 	imes 120} = \frac{184.8}{3125} = \frac{1848}{31250} = \frac{924}{15625}$. Wait,re-evaluating the expansion: the term $T_6$ involves $(n-4) = 0.2$ (positive) and $(n-5) = -0.8$ (negative). Thus,the first negative coefficient is the coefficient of $x^6$. Coefficient of $x^6 = \frac{\frac{21}{5} \cdot \frac{16}{5} \cdot \frac{11}{5} \cdot \frac{6}{5} \cdot \frac{1}{5} \cdot (-\frac{4}{5})}{6!} = \frac{21 \cdot 16 \cdot 11 \cdot 6 \cdot 1 \cdot (-4)}{5^6 \cdot 720} = \frac{-22176}{15625 \cdot 720} = \frac{-616}{15625 \cdot 20} = \frac{-616}{312500} = \frac{-616}{5^7}$.

The first negative coefficient in the terms occurring in the expansion of $(1+x)^{\frac{21}{5}}$ is

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Assertion $(A)$: If $|x| < 1$,then $\sum_{n=0}^{\infty}(-1)^n x^{n+1} = \frac{x}{x+1}$.
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?

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