If the number of terms in x-1- 1 x n n- I - is 401- then n is greater than

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Question

If the number of terms in ${(x + 1 + \frac{1}{x})}^{n} (n \in I^{+})$ is 401, then $n$ is greater than

201

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200

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199

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none of these

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Solution

The correct option is C 199
If we substitute 2 in place of 1 in the above expression, we get
$(x + 2 + \frac{1}{x})^{n}$
$((\sqrt{x} + \frac{1}{\sqrt{x}}))^{2 n}$
Since the total number of terms is 401
$2 n + 1 = 401$
$2 n = 400$
$n = 200$
However the actual question is
$(x + 1 + \frac{1}{x})^{n}$
Hence n is greater than 199.

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If the number of terms in (x+1+1x)n(n∈I+) is 401, then n is greater than

The correct option is C 199If we substitute 2 in place of 1 in the above expression, we get(x+2+1x)n((√x+1√x))2nSince the total number of terms is 4012n+1=4012n=400n=200However the actual question is(x+1+1x)nHence n is greater than 199.

If the number of terms in ${(x + 1 + \frac{1}{x})}^{n} (n \in I^{+})$ is 401, then $n$ is greater than