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

The coefficient of $t^{12}$ in the expansion of $(1 + t^2)^6(1 + t^6)(1 + t^{12})$ is:

The sum of all those terms which are rational numbers in the expansion of $(2^{1/3} + 3^{1/4})^{12}$ is:

If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^3} - x^4\right)^n, x \neq 0$, is zero, then the value of $n$ is . . . . . . .

The term independent of $x$ in the expansion of $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$ is

If the sum of the coefficients of $x^r$ $(r=0, 1, 2, \ldots, 2n)$ in the expansion of $(1+3x-2x^2)^n$ is $128$,then $\sum_{r=1}^{2n} r \frac{^{2n}C_r}{^{2n}C_{r-1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module