Question

If there is a term containing \(x^{2r}\) in \(\left(x + \frac{1}{x^2}\right)^{n-3}\), then

• A. \(n - 2r\) is a positive integral multiple of 3.
• B. \(n - 2r\) is even
• C. \(n - 2r\) is odd
• D. None of the above

Hint:

In binomial expansion, equate the exponent of \(x\) to the desired power.

Answer 1

The Correct Option is A

To determine if there is a term containing \(x^{2r}\) in the expansion of \(\left(x + \frac{1}{x^2}\right)^{n-3}\), we need to understand the binomial expansion of this expression.

The general term in the expansion of \((a + b)^m\) is given by: 

\(T_k = \binom{m}{k} a^{m-k} b^k\)

Here, \(a = x\) and \(b = \frac{1}{x^2}\), and \(m = n-3\), so our expression becomes \((x + \frac{1}{x^2})^{n-3}\).

The general term \(T_k\) in the expansion is:

\(T_k = \binom{n-3}{k} x^{n-3-k} \left(\frac{1}{x^2}\right)^k = \binom{n-3}{k} x^{n-3-k - 2k} = \binom{n-3}{k} x^{n-3-3k}\)

This implies that the power of \(x\) in the general term is \((n-3-3k)\).

We are given that the term contains \(x^{2r}\), thus we set:

\(n - 3 - 3k = 2r\)

Simplifying for \(n-2r\):

\(n - 3k = 2r + 3 \quad \Rightarrow \quad n - 2r = 3k\\)

This shows that \(n - 2r\) is a multiple of 3. Specifically, it is a positive integral multiple since both \(k\) and \(r\) are non-negative integers, which means the term exists with positive integers.

Thus, the correct answer is:

\(n - 2r\) is a positive integral multiple of 3.