Question

18．若函數(shù)f（x）=${C}_{n}^{0}$xⁿ+${C}_{n}^{1}{x}^{n-2}$+${C}_{n}^{2}{x}^{n-4}$+…+${C}_{n}^{r}{x}^{n-2r}$+…+${C}_{n}^{n}（\frac{1}{x}）^{n}$，其中n∈Nⁿ，則f′（1）=0．

Answer 1

分析 求出通項公式的導(dǎo)數(shù)，利用組合數(shù)的性質(zhì)，求解導(dǎo)函數(shù)值即可．

解答 解：$（{C}_{n}^{r}{x}^{n-2r}）′$=（n-2r）${C}_{n}^{r}{x}^{n-2r-1}$，
x=1時，上式=（n-2r）${C}_{n}^{r}$，
函數(shù)f（x）=${C}_{n}^{0}$xⁿ+${C}_{n}^{1}{x}^{n-2}$+${C}_{n}^{2}{x}^{n-4}$+…+${C}_{n}^{r}{x}^{n-2r}$+…+${C}_{n}^{n}（\frac{1}{x}）^{n}$，其中n∈N⁺，
f′（x）=n${C}_{n}^{0}$x^n-1+（n-2）${C}_{n}^{1}$x^n-3+（n-4）${C}_{n}^{2}$x^n-5+…+（n-2r）${C}_{n}^{r}$x^n-2t-1+…+（-n）${C}_{n}^{n}$x^-n-1．
則f′（1）=n${C}_{n}^{0}$+（n-2）${C}_{n}^{1}$+（n-4）${C}_{n}^{2}$+…+（n-2r）${C}_{n}^{r}$+…+（-n+2）${C}_{n}^{n-1}$+（-n）${C}_{n}^{n}$．
∵${C}_{n}^{0}={C}_{n}^{n}$，${C}_{n}^{1}={C}_{n}^{n-1}$，…
∴f′（1）=0．
故答案為：0．

點評 本題考查組合數(shù)的性質(zhì)的應(yīng)用，導(dǎo)數(shù)的應(yīng)用，考查計算能力．