If the last term in the binomial expansion of ${\left( {{2^{1/3}} - \frac{1}{{\sqrt 2 }}} \right)^n}$ is ${\left( {\frac{1}{{{3^{5/3}}}}} \right)^{{{\log }_3}8}}$, then the 5th term from the beginning is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Analyzing +4 | -1

If the last term in the binomial expansion of

{\left( {{2^{1/3}} - \frac{1}{{\sqrt 2 }}} \right)^n}

{\left( {\frac{1}{{{3^{5/3}}}}} \right)^{{{\log }_3}8}}

, then the 5th term from the beginning is

View answer and explanation

Correct option: A

210

Option A (210) is correct because after finding n=10, the 5th term can be calculated using the binomial theorem.

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\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n