2的n减一次方分之n的前n项和为

裂项an=(n+2)/[n!+(n+1)!+(n+2)!]=（n+2)[n!(1+n+1+(n+1)(n+2))]=（n+2)/[n!(n+2)^2]=1/[n!(n+2)]=(n+1)/(n+2)!

Sn=1/2+2/2^2+3/2^3+……+n/2^nSn*1/2=1/2^2+2/2^3+……+(n-1)/2^n+n/2^(n+1)上面两式相减Sn-Sn/2=1/2+1/2^2+1/2^3+……

运用错位相减法：an=n/2^nSn=a1+a2+a3+……+an=1/2^1+2/2^2+3/2^3+……+n/2^n(1/2)Sn=1/2a1+1/2a2+……+1/2an=1/2^2+2/2^3

答案为ASn=((a1+an)/2)*nan=a1+(n-1)d根据上式得出：Sn=(2a1+(n-1)d)*n/2=a1*n+n方*d/2-n*d/2limSn/n方=lim(2a1*n+n方*d-

M=1＋2＋3＋…＋n=[n(n＋1)]/2N=1²＋2²＋3²＋…＋n²=[n(n＋1)(2n＋1)]/6P=1³＋2³＋3³＋

不是n^2分之n,应该是n/2^nSn=1/2+2/2^2+3/2^3...+n/2^nSn/2=1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1)Sn-Sn/2=Sn/2=1/2

首先先把底都化为3式子就变为3^2n*3^(3n-3)/3^(3n+1)=3^4化简3^（2n-4）=3^4对比2n-4=4n=4则4^(-2)=1/16

Sn=25n-n^2a1=S1=25*1-1^2=24n≥2时an=Sn-S(n-1)=25n-n^2-[25(n-1)-(n-1)^2]=26-2n令an=0得n=13所以当n≤13时Tn=|a1|

Sn=1+3/2+5/2^2+...+(2n-1)/2^(n-1)(1/2)Sn=1/2+3/2^2+5/2^3+...+(2n-3)/2^(n-1)+(2n-1)/2^n错位相减法得(1-1/2)S

Sn=2^n-1---------(1)当n=1时,a1=1S(n-1)=2^(n-1)-1-------(2)(1)-(2)Sn-S(n-1)=2^n-2^(n-1)an=2^(n-1)a1+a3+

A1=6;n>1时,an=Sn-S(n-1)=6n-1.

n=1时,a1=S1=4×1²+2×1=6n≥2时,an=Sn-S(n-1)=4n²+2n-[4(n-1)²+2(n-1)]=8n-2n=1时,a1=8×1-2=6,同样

数列为:an=(2n-1)/2^n2sn=1+3/2+5/4+7/8+9/16+...+(2n-1)/2^n-1sn=2sn-sn=1+2(1/2+1/4+1/8+...+1/2^n-1)-(2n-1

已知等差数列{an}的公差为2,其前n项和Sn=pn²+2n（n∈N*）．（I）求p的值及an；（II）若bn=2/﹙2n-1﹚an,记数列{bn}的前n项和为Tn,求使Tn﹥9/10成立的

S=0.25n(n+1)(n+2)(n+3)再问：能提供方法么？谢谢！是用裂项么？再答：n(n+1)(n+2)=0.25[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

2^(4n+1)-4^(2n-1)+16^n=2*2^4n-4^2n÷4+4^2n=2*4^2n-4^2n÷4+4^2n=(2-1/4+1)*4^2n=(11/4)*4^2n=11*4^(2n-1)

前n项偶数和前n项奇数共2n项,则（2n）^3-n^2(4n+3)即可.

2^n-2

an=sn-sn-1=n^2+3n-(n-1)^2-3(n-1)=2n-1+3=2(n+1)an-an-1=2(n+1)-2n=2所以为等差数列