Discussion
If n is a positive integer and the coefficient of x in the expansion of \(\left(x^{2} + \frac{1}{x^{3}}\right)^{n}\) is \({}^{n}C_{7}\), then n is equal to
Answer: A
\(n = 18\).
The general term of \(\left(x^{2} + \frac{1}{x^{3}}\right)^{n}\) is \({}^{n}C_{r}\, x^{2(n-r)}x^{-3r} = {}^{n}C_{r}\, x^{2n - 5r}\). For the term in x, \(2n - 5r = 1\).
The coefficient is \({}^{n}C_{7}\), so \(r = 7\) (or \(r = n - 7\)). With \(r = 7\): \(2n = 36\), so \(n = 18\). (The alternative \(r = n - 7\) gives \(2n - 5n + 35 = 1\), \(n = \frac{34}{3}\), not an integer.)
Check: for \(n = 18\), \(r = 7\) gives the power \(36 - 35 = 1\). The other values of n leave no integer r with coefficient \({}^{n}C_{7}\).
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