已知数列的前20项的和为30,前30项的和为70
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因为:(5n-8)Sn+1-(5n+2)Sn=-20n-8...(1)所以:(5(n+1)-8)Sn+2-(5(n+1)+2)Sn+1=-20(n+1)-8即:(5n-3)Sn+2-(5n+7)Sn+
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+..
设此数列的前10项和为x,由已知得:a1+a2+……+a5=3,a6+a7+……+a10=x-3,a11+a12+……+a15=39-x.对于等比数列来说,a1+a2+……+a5,a6+a7+……+a
(1)当n=1时,a1=S1=9,当n≥2时,an=Sn-Sn-1=10n-n2-[10(n-1)-(n-1)2]=11-2n,当n=1时,a1=9,满足an=11-2n,所以an=11-2n,n∈N
因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易
X(),括号里的是下标,^是次方logX(n+1)-logX(n)=1log(X(n+1)/X(n))=1=log10X(n+1)/X(n)=10{Xn}是公比为10的等比数列S100=X1(10^1
S(n-1)=2a(n-1)-1所以Sn-S(n-1)=2an-2a(n-1)因为Sn-S(n-1)=an所以an=2an-2a(n-1)所以an=2a(n-1)an/[a(n-1]=2所以an是等比
#includemain(){inta1=1,a2=2,a3=3,a;inti;for(i=0;i
证:(1)根号Sn+1=(a1+1)*2^(n-1)=4*2^(n-1)=2^(n+1)Sn+1=2^(2n+2)=4^(n+1).1Sn=4^n.21式-2式Sn+1-Sn=4^(n+1)-4^na
Sn=(n^2+n)/21/Sn=1/((n2+n)/2)=2/(n^2+n)Tn=1+2/6+2/12+2/30+.+2/n*(n+1)=1+(2/2-2/3)+(2/3+2/4)+.+(2/n-2
(1)当n=1时,a1=S1=lg1=0(2)当n>=2时,an=Sn-Sn-1=lgn-lg(n-1)=lg(n/(n-1))所以,a1=0;an=lg(n/(n-1))(n>=2)(可把结果写成分
AnA(n+1)在分子上?.是的话S1=A1=1=1/2(A1A2)=1/2(A2)A2=2S(n+1)=Sn+A(n+1)=1/2(AnA(n+1))+A(n+1)=1/2(A(n+1)A(n+2)
是等差数列,所以前十五项的平均是中间那个数,即第八项所以第八项等于135/15=9第八项为9
因为:数列{log2Xn}是公差为1的等差数列所以:log2xn-log2x(n-1)=log2[xn/x(n-1)]=1则:xn/x(n-1)=2所以:{xn}成等比数列S100=100S200=a
Xn+1/Xn=2答案为100(2^100+1)
解题思路:方法:数列通项的求法:已知sn,求an。求和:错位相减法。解题过程:
an=要分段8-2n^2,当n=1时2-2n,当n大于1时
∵Sn=kq^n-k∴S(n+1)=kq^(n+1)-k∴a(n+1)=S(n+1)-Sn=[kq^(n+1)-k]-(kq^n-k)=k[q^(n+1)-q^n]=k[(q-1)q^na(n+1)/
当n=1时,a1=S1=1当n≥2时,an=Sn-S(n-1)=3n²-2n-3(n-1)²+2(n-1)=6n-5∵当n=1时,满足an=6n-5又∵an-a(n-1)=6n-5
Sn-S(n-1)=2An-2A(n-1)=An所以An=2A(n-1)An/2A(n-1)=2即An为等比为2的等比数列令n=1,S1=3+2A1=A1A1=-3所以An=-3*[2^(n-1)]