求前n项之和1-2 3 3 5-4 7
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/05 06:57:32
(1)该数列通项an=1+2+3+.+n=1/2*n^2-1/2*n令bn=1/2*n^2,cn=1/2*n,则an=bn+cn数列{bn}的前n项和sn'=1/2*(1^2+2^2+3^2+.+n^
1a1q^2*a1q^7=1a1^2*q^9=1Sn=a1(1-q^n)/(1-q)Tn=(1/a1)(1-1/q^n)/(1-1/q)Tn<Sn得q/(a1q^n)<a1a1>0时q<a1q^nq<
哇,乘公比错位相减:(费劲)
因为an=(5/9)×((10^n)-1))所以sn=(5/9)×((10^1)-1))+(5/9)×((10^2-1))+…………+(5/9)×((10^n)-1))=(5/9)×[(10^1+10
an=(5/9)(10^n-1)接下来等比数列求和an=(5/9)10^n-(5/9)bn=an+(5/9)=(5/9)10^nSbn=(50*10^n-50)/81=San+(5/9)nSan=(5
Sn=1+2+3+---+n+1/2+1/4+1/8+----+1/2^n=n(n+1)/2+(1/2)[1-(1/2)^n]/[1-(1/2)]=n(n+1)/2+1-(1/2)^n=[n^2+n+
错位相减法Sn=1×2+3×2²+5×2³+.+(2n-1)×2ⁿ①2Sn=2²+3×2³+5×2⁴+.+(2n-3)×2ⁿ
#includeintmain(){inti=1,j=1,n;floatsum=0;scanf("%d",&n);for(intk=0;k
#include<stdio.h>int main(){ int i,j,n,s=0;
a1=5an=-3^n/2+2(n≥2)
A15=A10*q^5A10=q^5A1=q^(-5)An=q^(n-6)Sn=(1-q^n)/((1-q)*q^5)Bn=q^(6-n)倒数数列首项Tn=(1-q^n)/(1-q)*q^(6-n)自
#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]
Sn=(103n-3n^2)/2S1=a1=50Sn-1=[103(n-1)-3(n-1)^2]/2Sn-Sn-1=an=53-3na1a2……a17都是正数,后面的是负数设Tn=|an|的n项之和n
设等比数列首项为a,比为q则Sn=a(1-q^n)/(1-q)倒数的数列首项为1/a,比为1/qTn={1/a[(1-(1/q)^n)]}/(1-1/q)=q(q^n-1)/[aq^n(q-1)]Sn
设A(n)=1+3×2+5×2²+7×2³+...+(2n-1)*2^(n-1)2A(n)=2+3×2²+5×2³+...+(2n-3)*2^(n-1)+(2n
An=4n-1所以a1=3a10=40-1=39所以s10=(a1+a10)×10÷2=(3+39)×5=42×5=210
n=(-1)^n*n²Tn=-1²+2²-3²+……+(-1)^n*n²则若n是奇数平方差,2-1=4-3=……=1Tn=(2+1)+(4+3)+……
错位相减法,Sn=(-1)^n*(3n/2-1/4)-1/4
an=n+(1/2)^n;Sn=1+1/2+2+1/4+3+1/8+.+n+(1/2)^n;=(1+2+3+.+n)+(1/2+1/4+1/8+.(1/2)^n);=n(n+1)/2+(1-(1/2)