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

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(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

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$6$

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(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 $....$ .

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