题目内容
18.先阅读下面的推理过程,然后完成下面问题:在等式cos2x=2cos2x-1(x∈R)的两边对x求导,即(cos2x)′=(2cos2x-1)′;
由求导法则得(-sin2x)•2=4cosx•(-sinx)化简后得等式sin2x=2sinxcosx.
(Ⅰ)已知等式(1+x)n=${C}_{n}^{0}$+${C}_{n}^{1}$x+${C}_{n}^{2}$x2+…+${C}_{n}^{n-1}$xn-1+${C}_{n}^{n}$xn(x∈R,整数n≥2),证明:n[(1+x)n-1-1]=$\sum_{k=2}^{n}$k${C}_{n}^{k}$xk-1;
(Ⅱ)设n∈N*,x∈R,已知(2+x)n=a0+a1x+a2x2+…+anxn,令bn=$\frac{n({n}^{2}+1)({a}_{0}-{2}^{n-1})}{{a}_{1}+2{a}_{2}+3{a}_{3}+…+n{a}_{n}}$,求数列{bn}的最大项.
分析 (Ⅰ)对等式(1+x)n=${C}_{n}^{0}$+${C}_{n}^{1}$x+${C}_{n}^{2}$x2+…+${C}_{n}^{n-1}$xn-1+${C}_{n}^{n}$xn的x求导,整理即可得出结论;
(Ⅱ)先求出a0的值,再对等式中的x求导,利用特殊值求出bn的通项公式,从而求出数列{bn}的最大项.
解答 解:(Ⅰ)证明:在等式(1+x)n=${C}_{n}^{0}$+${C}_{n}^{1}$x+${C}_{n}^{2}$x2+…+${C}_{n}^{n-1}$xn-1+${C}_{n}^{n}$xn(x∈R,整数n≥2)
的两边对x求导,得:
n(1+x)n-1=${C}_{n}^{1}$+2${C}_{n}^{2}$x+…+(n-1)${C}_{n}^{n-1}$xn-2+n${C}_{n}^{n}$xn-1,
移项,得:n[(1+x)n-1-1]=$\sum_{k=2}^{n}$k${C}_{n}^{k}$xk-1;
(Ⅱ)由等式(2+x)n=a0+a1x+a2x2+…+anxn,
得a0=2n,
再两边对x求导,得:
n(2+x)n-1=a1+2a2x+…+nanxn-1;
令x=1,得:a1+2a2+3a3+…+nan=n×3n-1;
则bn=$\frac{n({n}^{2}+1)({a}_{0}-{2}^{n-1})}{{a}_{1}+2{a}_{2}+3{a}_{3}+…+n{a}_{n}}$=$\frac{n{(n}^{2}+1){×2}^{n-1}}{n{×3}^{n-1}}$=(n2+1)×${(\frac{2}{3})}^{n-1}$;
又bn+1-bn=[(n+1)2+1]×${(\frac{2}{3})}^{n}$-(n2+1)×${(\frac{2}{3})}^{n-1}$
=${(\frac{2}{3})}^{n-1}$×$\frac{{-n}^{2}+4n+1}{3}$,
得:n≤4时,bn+1>bn,n≥5时,bn+1<bn;
所以数列{bn}的最大项为b5=${(\frac{2}{3})}^{4}$×26=$\frac{416}{81}$.
点评 本题考查了类比推理的应用问题,也考查了导数的综合应用问题,考查了等比数列的求和公式的应用问题,是综合性题目.
A. | -1 | B. | 0 | C. | 1 | D. | 2 |
气温/(℃) | 4 | 2 | 1 | -1 | -3 |
杯数 | 24 | 36 | 40 | 49 | 61 |
A. | y=4x+36 | B. | y=5x+20 | C. | y=-4x+44 | D. | y=-5x+45 |