(x^n-1) (x-1) 计算增长-搜答案-灵鹊做题网

importjava.util.*;publicclassTest{\x05\x05publicstaticvoidmain(Stringargc[]){\x05\x05Scannercin=newS

f(x)=1*x^0+2*x^1+.+n*x^(n-1)+.xf(x)=x+...+(n-1)x^(n-1)+nx^n+...f(x)-xf(x)=1+x+x^2+x^3+...x^(n-1)-nx^

#include#includeintmain(){doublex1,x2,a;printf("pleaseinputanumble:");scanf("%lf",&a);if(a=1e-5);pri

PrivateSubCommand1_Click()DimnAsLong,xAsSingle,yAsSingle,y2AsSingle,sumAsSinglex=Val(InputBox("请输入一个

n

下面是M文件代码,参数需要调整x=0.5;n=10;%只能是正整数s=1;fori=1:ns=s+x^i/factorial(i);%factorial(n)计算n的阶乘end

doubley=k=s=t=1;这个分开来写：doubley=1,k=1,s=1,t=1;

(1)(-x)^2n×(-2^2n)×(-x)^2n+1,2n为正整数=x^2n*(-2^2n)*(-x^2n+1)=2^(2n)*x^(4n+1)2)(2a-b)^4×(b-2a)^3×(2a-b)

（1/k!)X[1/(n-k)!]=1/[k!X(n-k)!]=1/(n!Xk)

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

1、原式=6x*2ny²×2x*（n-1）=12x*（3n-1）y³2、底面周长=2x×3.14=6.28x表面积=6.28x×（2x-5）+2×3.14x²=6.28x

longfun(intx,intn)//这个函数计算x的n次方{inti;longq=1;if(n==0){return1;}if(n==1){returnx;}for(i=1;i

(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)/

阶乘函数,不能直接修改m的值.不过此处无影响.主要是没加1.#include#include#includeintfact(intm){intt=1;inti;//if(m==0)//return1;

(1)(m-n)^2*(n-m)^5*(m-n)^5=(m-n)^2*[-(m-n)^5]*(m-n)^5=-(m-n)^(2+5+5)=-(m-n)^12(2)(-x)^2*x^3+2x^3*(-x

（x^3)^n÷x^2n÷x^n-1=（x^3n）÷x^2n÷x^n-1=x^（3n-2n）÷x^n-1=x^n÷x^n-1=x^(n-n+1)=x

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&

=-8（-x^(n-1))^3=-8(-x^3(n-1))=8x^(3n-3)