The first negative coefficient in the terms o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

For $n, p \in N-\{1\}$,the coefficient of $x^3$ in $\frac{(1-x)^{-1 / p}}{(1-x)^n}$ is:

The sum of the infinite series $1+\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots$ is equal to

The coefficient of $x^2$ in the expansion of $(1-3x)^{\frac{1}{3}}(1+2x)^{-\frac{1}{2}}$ is

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ is:

The coefficient of ${x^n}$ in the expansion of ${(1 - 9x + 20{x^2})^{-1}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module