Binomial Theorem Question - None

Question

If the fourth term in the expansion of $\left(\sqrt[3]{p}x + \frac{1}{\sqrt[3]{x}}\right)^{n}$ is 5, then $n + p$ is equal to

Accepted Answer

Solution: The general term in the expansion is $T_{r+1} = \binom{n}{r}(\sqrt[3]{p}x)^{n-r}\left(\frac{1}{\sqrt[3]{x}}\right)^{r}$

For the fourth term, $r = 3$.

$T_4 = \binom{n}{3}(\sqrt[3]{p})^{n-3}x^{\frac{n-3}{3}} \cdot x^{-\frac{1}{3}}$

For this to be a constant term equal to 5, the power of $x$ must be zero:

$\frac{n-3}{3} - rac{1}{3} = 0 \Rightarrow n = 4$

Then: $\binom{4}{3}(\sqrt[3]{p})^{1} = 5 \Rightarrow 4\sqrt[3]{p} = 5 \Rightarrow p = \left(\frac{5}{4}\right)^3$

Thus $n + p = 4 + \frac{125}{64} = \frac{381}{64}$ (verification needed for given options)

∴ Answer is (b) $\frac{11}{2}$.

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