If the last term in the binomial expansion of 21/3−12n is 135/3log3⁡8, the the 5th term from  the begging is

# If the last term in the binomial expansion of , the the 5th term from  the begging is

1. A

210

2. B

420

3. C

105

4. D

None of these

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### Solution:

$\begin{array}{r}{\mathrm{T}}_{\mathrm{n}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\left({2}^{1/3}\right)}^{\mathrm{n}-\mathrm{n}}{\left(-\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}}\\ {=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\left(-1{\right)}^{\mathrm{n}}\frac{1}{{2}^{\mathrm{n}/2}}=\frac{\left(-1{\right)}^{\mathrm{n}}}{{2}^{\mathrm{n}/2}}\end{array}$

Also, we have

${\left(\frac{1}{{3}^{5/3}}\right)}^{{\mathrm{log}}_{3}8}={3}^{-\left(5/3\right){\mathrm{log}}_{3}{2}^{3}}={2}^{-5}$

Now $\begin{array}{r}{\mathrm{T}}_{5}={\mathrm{T}}_{4+1}{=}^{10}{\mathrm{C}}_{4}{\left({2}^{1/3}\right)}^{10-4}{\left(-\frac{1}{\sqrt{2}}\right)}^{4}\\ =\frac{10!}{4!6!}{\left({2}^{1/3}\right)}^{6}\left(-1{\right)}^{4}{\left({2}^{-1/2}\right)}^{4}\\ =210\left(2{\right)}^{2}\left(1\right)\left({2}^{-2}\right)=210\end{array}$

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