The smallest natural number n, such that the coefficient of x in the expansion of (x^2+1x)…

Question

The smallest natural number $n$, such that the coefficient of $x$ in the expansion of $\left(x^2+\dfrac1x\right)^n$ is ${}^nC_{23}$, is:

Accepted Answer

$$T_{r+1}=\binom nr x^{2(n-r)-r}=\binom nr x^{2n-3r}$$

For coefficient of $x$: $2n-3r=1$. Given the binomial coefficient equals ${}^nC_{23}$, so $r=23$ or $n-r=23$, giving $n=58$ or $n=38$ respectively (checking $2n-3r=1$ against each). Smallest value: $n=38$.

Solutions · 1