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Question: Given that the $4^{\text{th}}$ term in the expansion of $\left(2+ \frac{3x}{8}\right)^{11}$ has the maximum numerical value, find the range of values of $x$ for which this is true.


Answer: I am not sure of the answer since my answer doesn't match. Please clear my doubt. I wonder if my calculation is wrong or my concepts aren't clear. The answer is $-\frac{64}{21} < x < -2$ and $2< x < \frac{64}{21}$ my answer comes exactly with opposite signs.

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  • $\begingroup$ For $x>0$ we have $56<21x<80.$ $\endgroup$
    – Lucian
    Dec 12 '15 at 4:30
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One way of doing this:

The $r^{th}$ term of $(2+ \frac{3x}{8})^{11}$ is ${11\choose r} 2^{11-r} \left(\frac{3x}{8}\right)^r$.

Hence $$\frac{d}{dx} \left[{11\choose r} 2^{11-r} \left(\frac{3x}{8}\right)^r\right] = 0$$ at $r=4$.

Hope this helps.

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