Let a_0, a_1, ldots, a_{23} be real numbers s | vedclass

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

Question

(B) The general term of the expansion $(1 + \frac{2}{5}x)^{23}$ is given by $T_{r+1} = \binom{23}{r} (\frac{2}{5}x)^r$. The coefficient $a_r$ is $\bin

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The general term of the expansion $(1 + \frac{2}{5}x)^{23}$ is given by $T_{r+1} = \binom{23}{r} (\frac{2}{5}x)^r$. The coefficient $a_r$ is $\binom{23}{r} (\frac{2}{5})^r$. To find the largest coefficient $a_r$,we consider the ratio $\frac{a_r}{a_{r-1}} \geq 1$. $\frac{\binom{23}{r} (\frac{2}{5})^r}{\binom{23}{r-1} (\frac{2}{5})^{r-1}} \geq 1$ $\frac{23-r+1}{r} 	imes \frac{2}{5} \geq 1$ $\frac{24-r}{r} 	imes \frac{2}{5} \geq 1$ $48 - 2r \geq 5r$ $48 \geq 7r$ $r \leq \frac{48}{7} \approx 6.85$. Since $r$ must be an integer,the largest coefficient occurs at $r = 6$.

Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that $(1+\frac{2}{5} x)^{23} = \sum_{i=0}^{23} a_i x^i$ for every real number $x$. Let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$. Then the value of $r$ is $....$ .

Similar Questions

If the numerically greatest term in the expansion of $(2-3x)^9$ when $x=1$ is $P_1^\alpha P_2^\beta P_3^\gamma P_4^\delta$ (where $P_1 < P_2 < P_3 < P_4$ are the first four prime numbers),then $\alpha+\beta+\gamma+\delta=$

In the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$,the coefficient of $x^4$ is

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$,$n \in N$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

The middle term in the expansion of ${\left( {x - \frac{1}{x}} \right)^{18}}$ is

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0$, if the term independent of $x$ is $(221)k$, then $k$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module