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Proof by mathematical induction

Binomial expa...

Question

Binomial expansion of ${(x^{k} + \frac{1}{2^{2 k}})}^{3 n}$ , where $n$ is a positive integer, always contains a term which is independent of

x^{2}

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x

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x^{3}

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none of the above

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Solution

The correct option is A $x$
In the above question
$T_{r + 1} =^{3 n} C_{r} x^{3 k n - k r} 2^{- 2 k r}$
For a term independent of $x$
$3 k n - k r = 0$
$k (3 n - r) = 0$
Hence $k = 0$ or $r = 3 n$

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