sn=2的n次方-1,求奇数项的和

设通项为anSn-Sn-1=an=2^(n-1)(n≥2)又S1=a1=1符合条件,故an=2^(n-1)(n∈N*)于是奇数项的前n项和NN=a1+a3+...+a2n-1=1+2^2+2^4+..

an=(2n-1)(1/4)^n=n(1/4)^(n-1)-(1/4)^nSn=a1+a2+..+an=[summation(i:1->n){i(1/4)^(i-1)}]-(1/3)(1-(1/4)^

a1+a2=2*1-1+3*2+2=9a(n-1)+an=2*(2-1)-1+3n+2=5n-1等差a9+a10=49S10=(9+49)*5/2=145S15=S14+a15Sn同理可求

由题意得a(n+1)=Sn+1-Sn=Sn+3^n即Sn+1=2Sn+3^n整理得Sn+1-3^（n+1）=2(Sn-3^n)设Sn-3^n=bn则｛bn｝是以b1为首项,2为公比的等比数列b1=S1

n=(n-1)/2^(n-1)Sn=b1+b2+...+bn=(1-1)/2^0+(2-1)/2^1+(3-1)/2^2+...+(n-1)/2^(n-1)=1/2^1+2/2^2+...+(n-1)

n=1时,a1=S1=2a1-2²a1=4n≥2时,Sn=2an-2^(n+1)S(n-1)=2a(n-1)-2ⁿSn-S(n-1)=an=2an-2^(n+1)-2a(n-1)

此为等差数列求和+等比数列求和若n为偶数等差数列首项为5,公差为4等比数列首项为16,公比为16Sn=[5+2（n-1)+3]*(n/2)/2+16[1-16^(n/2)]/(1-16)若n为奇数则将

n为奇数项,an-a(n-2)=2n-2(n-2)=4,奇数项为首项为2,公差为4的等差数列.n为偶数项,an/a(n-2)=2^n/2^(n-2)=4,偶数项为首项为4,公比为4的等比数列.n为偶数

s(n-1)=2a(n-1)+(-1)^(n-1)这两个作差an=2an-2a(n-1)+(-1)^n-(-1)^(n-1)得an=2a(n-1)-(-1)^n+(-1)^(n-1)两边同除以2^n；

1.当n为偶数时,n=2ka(2k-1)=6(2k-1)-5)=12k-11sk=12k(k+1)/2-11k=6k^2-5ka(2k)=2^(2k)=4^ktk=4(4^k-1)/3=(1/3)4^

解Sn=2n²-3nS(n-1)=2(n-1)²-3(n-1)(n≥2)an=Sn-S(n-1)=2n²-3n-2(n-1)²+3(n-1)=4n-5当n=1时

：（I）当n=1时,a1=S1==2,当n≥2时,an=Sn-Sn-1=（n2+3n-（n-1）2-3（n-1）=n+1,∴an=n+1（n）,当n为偶数时,Tn=（a1+a3+…+an）+（22+2

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+

采用Sn-q倍Sn,错位相减法!an=(2n-1)*(1/2)^nSn=1*(1/2)+3*(1/2)^2+5*(1/2)^3+……+(2n-1)*(1/2)^n0.5Sn=1*(1/2)^2+3*(

分组求和Sn=a1+a2+a3+……+an=(1+1/2)+(3+1/4)+(5+1/8)+……+[(2n-1)+1/2^n]=(1+3+5+……+(2n-1))+(1/2+1/4+1/8+……+1/

ifn奇数Sn=(n+1)(an+a1)/4+(n-1)((a2+a(n-1))/4=(n+1)(2n-2)/4+(n-1)(5+2(n-1)+1)/4=(n+1)(2n-2)/4+(n-1)(2n-

若n=2kSn=(4+3(2k-1)+1)/2+2^k-2=2^k+3k-1=2^(n/2)+3n/2-1若n=2k+1Sn=2^k+3k-1+3(2k+1)+1=2^k+9k+3=2^((n-1)/

Sn+1+Sn-1=2Sn+1(Sn+1-Sn)+(Sn-1-Sn)=1(an+1)-an=1so等差数列接下来能做了吧你

1.当n为偶数时偶数项和和奇数项各有n/2项；奇数项为等差数列,a1=1,尾项为a(n-1)=6n-11各项和S奇＝[a1+a(n-1)]*(n/2)/2=3n(n-2)/2偶数项为等比数列,a2=1

a(n)=n*2^n,S(n)=a(1)+a(2)+a(3)+...+a(n-1)+a(n)=1*2+2*2^2+3*2^3+...+(n-1)*2^(n-1)+n*2^n,2S(n)=1*2^2+2