Question

The natural number \(m\), for which the coefficient of \(x\) in the binomial expansion of \(\left(x^{m}+\frac{1}{x^{2}}\right)^{22}\) is 1540, is ............. .

Hint:

The coefficient is \(\binom{22}{r}\) for some \(r\); \(\binom{22}{r}=1540\) forces \(r=3\) or \(19\), then match the power of \(x\).

Answer 1

Correct Answer: 13

General term: \(T_{r+1}=\binom{22}{r}(x^m)^{22-r}\left(x^{-2}\right)^{r}=\binom{22}{r}x^{22m-mr-2r}\).
The power of \(x\) is 1: \(22m-r(m+2)=1\).
The coefficient is \(\binom{22}{r}=1540\). Since \(\binom{22}{3}=\binom{22}{19}=1540\), \(r=3\) or \(r=19\).
\(r=3\): \(22m-3m-6=1\Rightarrow19m=7\), not a natural number.
\(r=19\): \(22m-19m-38=1\Rightarrow3m=39\Rightarrow m=13\).
Answer: 13