First slide

Binomial theorem for positive integral Index

Question

If the last term in the binomial expansion of ${(2^{1 / 3} - \frac{1}{\sqrt{2}})}^{n} is {(\frac{1}{3^{5 / 3}})}^{\log_{3} 8}$ , the the 5th term from the begging is

Moderate

A

210
B

420
C

105
D

None of these

Solution

$Last term of {(2^{1 / 3} - \frac{1}{\sqrt{2}})}^{n} is$
$\begin{aligned} T_{n + 1} =^{n} C_{n} {(2^{1 / 3})}^{n - n} {(- \frac{1}{\sqrt{2}})}^{n} \\ =^{n} C_{n} (- 1)^{n} \frac{1}{2^{n / 2}} = \frac{(- 1)^{n}}{2^{n / 2}} \end{aligned}$
Also, we have
${(\frac{1}{3^{5 / 3}})}^{\log_{3} 8} = 3^{- (5 / 3) \log_{3} 2^{3}} = 2^{- 5}$
$\begin{aligned} Thus, \frac{(- 1)^{n}}{2^{n / 2}} = 2^{- 5} \\ \Rightarrow \frac{(- 1)^{n}}{2^{n / 2}} = \frac{(- 1)^{10}}{2^{5}} \\ \Rightarrow \frac{n}{2} = 5 \Rightarrow n = 10 \end{aligned}$
Now $\begin{aligned} T_{5} = T_{4 + 1} =^{10} C_{4} {(2^{1 / 3})}^{10 - 4} {(- \frac{1}{\sqrt{2}})}^{4} \\ = \frac{10!}{4! 6!} {(2^{1 / 3})}^{6} (- 1)^{4} {(2^{- 1 / 2})}^{4} \\ = 210 (2)^{2} (1) (2^{- 2}) = 210 \end{aligned}$

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