(x-1|x)^2n-搜答案-搜搜考试网

若x=1Sn=1+2+3+……+n=n(n+1)/2若x不等于1xSn=x+2x^2+3x^3+……+n*x^n所以Sn-x*Sn=1+x+x^2+x^3+……+x^(n-1)-n*x^nSn(1-x

f(x)为n+1阶多项式,所以n+1阶求导后只会剩下x的n+1次方的导数,为n+1的阶乘

[x^n-x^(n-1)+x]*x^(n+1)=x^(2n+1)-x^2n+x^(n+2)

Sn是等差数列an=2n-1,等比数列bn=(-x)^(n-1)前n+1的和Sn-(-x)Sn=(a1b1+a2b2+...an+1bn+1)-(a1b2+a2b3+...+an+1bn+2)=a1b

答案是(x-1)^2注意,2x^3y指的是2*x^3*y;你的意思,正确的写法是2x^(3y)再问：额就写答案鬼知道你步骤是什么？再答：这个题目直接：x^(2n)*x^(2-2n)-x^(2n)*x^

#include<stdio.h>#include<math.h>//f(x)=1+x+x^2/2!+x^3/3!+...+x^n/n!直到|x^n/n|<10^-6do

原式=x^(5n)-x^(4n)+x^(2n)+x^(4n)-x^(3n)+x^n+x^(3n)-x^(2n)+1=x^(5n)+x^n+1

(-x)^2n十1.(-x)^n+1=（-x）^3n+2=-x^3n+2

(m+2n)/(n-m)+n/(m-n)-2m(n-m)=(m+2n-n-2m)/(n-m)=(n-m)/(n-m)=1[(x+2)/(x×x-2x)-(x-1)/(x×x-4x+4)]÷(x-4)/

M是指任意正实数.|x(n)|

4x^(n+2)-9x^n+6x^(n-1)-x^(n-2)=x^(n-2)(4x^4-9x^2+6x^2-1)=x^(n-2)[4x^4-(3x-1)²]=x^(n-2)(2x²

s(x)*(1+x)=1-x+x+x^2-x^2-……=1s(x)=1/(1+x)

lim(x->1)(x^(n+1)-(n+1)x+n)/(x-1)^2=lim(x->1)(x^(n+1)-(n+1)x+n)'/((x-1)^2)'=lim(x->1)((n+1)x^n-(n+1)

x^3n/(x^n-1)-x^2n/(x^n+1)-1/(x^n-1)+1/(x^n+1)=(x³ⁿ-1)/(xⁿ-1)-(x²ⁿ-1)/(x&

x/2+x/6+x/12+x/20+...+x/(n-1)n>n-1x×[1/2+1/6+1/12+1/20+……+1/(n-1)n]＞n-1x×[1-1/2+1/2-1/3+1/3-1/4+1/4-

x^n*x^n+1+x^2n*x=x^(2n+1)+x^(2n+1)=2x^(2n+1)

证明：当n=1,2时,两式相等,当n>2时,用二项式展开,提出其中的相同项,即x^n+(1/x)^n则只需比较剩下部分,用数学归纳法就可证明了.