数列an满足a1=1a2=√5,当n≥2时,
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an=1+2+3+…+n=[n(n+1)]/2则:1/(an)=2/[n(n+1)]=2[(1/n)-1/(n+1)],所以:M=1/(a1)+1/(a2)+1/(a3)+…+1/(an)=2[1/1
由an+2=3an+1-2an可得an+2-an+1=2(an+1-an)因为a2-a1=2,所以an+1-an不会等于0,则an+1-an是以2为公比的等比数列由上可得an+1-an=2^nan-a
1+2+3+.+n-1=(1+n-1)(n-1)/2等差数列求和哦~所以跟外面的2约了!
1,1/a1+2/a2+3/a3+…+n/an=2n那么1/a1+2/a2+3/a3+…+(n-1)/a(n-1)=2(n-1)两式相减,得:n/an=2n-2(n-1)=2那么an=n/22,Sn=
(1)证b1=a2-a1=1,当n≥2时,bn=an+1−an=an−1+an2−an=−12(an−an−1)=−12bn−1,所以{bn}是以1为首项,−12为公比的等比数列.(2)解由(1)知b
(1){an}是等差数列,a1=1,a2=a(a>0),an=1+(n-1)(a-1)a3=2a-1,a4=3a-2b3=a3*a4=(2a-1)(3a-2)=12a=2,或-5/6(舍去)所以a=2
设前n项和为Sn,Sn=n的平方,那么前(n-1)项S(n-1)的和为(n-1)的平方.Sn-S(n-1)=an{an}的通项就是n的平方减(n-1)的平方结果是2n-1哎呀我的妈呀不会打n的平方累死
n=√an*a(n+1)b(n+1)=√a(n+1)a(n+2)[b(n+1)/bn]^2=[a(n+1)*a(n+2)]/[a(n+1)*an]=a(n+2)/ana(n+2)=q^2*an
令Sn=a1/1+a2/3+a3/5+…+an/(2n-1)=3^(n+1),则:an=Sn-S(n-1)其中n>1,n∈N+∴an/(2n-1)=3^(n+1)-3^(n)=2·3^(n)其中n>1
A1=1/2成立,设An=1/[n(n+1)]成立,因为A1+A2+…+An=n^2An所以A1+A2+…+An+A(n+1)=(n+1)^2A(n+1),所以A(n+1)=(n+1)^2A(n+1)
∵数列{a[n]}满足a[1]+2a[2]+3a[3]+...+na[n]=(n+1)(n+2)∴a[1]+2a[2]+3a[3]+...+na[n]+(n+1)a[n+1]=(n+2)(n+3)将上
a1+a2+a3+...+an=n^2+2n可得:Sn=a1+a2+a3+...+an=n^2+2n当n=1时有:a1=S1=1+2=3当n≥2时有:an=Sn-S(n-1)=n^2+2n-(n-1)
据题意:5+(n-1)*d=5*(n-1)+(1+2+···n-2)*d5+(n-1)*d=5n-5+{[(n-2)(n-1)]/2}*d5+n*d-d=5n-5+[(n^2)/2]*d-(3n/2)
(I)∵bn=an+1-an,∴an+2-2an+1+an=bn+1-bn=2n-6∴bn−bn−1=2(n−1)−6,bn−1−bn−2=2(n−2)−6,…,b2−b1=2−6将这n-1个等式相加
a(n+2)=3*a(n+1)-2*ana(n+2)-a(n+1)=2*(a(n+1)-an)a2-a1=3-1=2a(n+1)-an=2^na(n+2)-2a(n+1)=a(n+1)-2*ana2-
a2=a1+2a2=1+2a2得a2=-1an=a1+2a2+3a3+...+(n-2)a(n-2)+(n-1)a(n-1)a(n-1)=a1+2a2+3a3+...+(n-2)a(n-2)两式相减:
(1)证明:由条件得a[n+2]-a[n+1]=2(a[n+1]-a[n])首项为a[2]-a[1]=5-2=3,公比为2,所以{a[n+1]-a[n]}为等比数列由(1)得a[n+1]-a[n]=3
a1=10an=9*10的n-1次方
lg(1+a1+a2+.+an)=n1+Sn=10^nSn=10^n-1n=1时,a1=S1=9n≥2时,an=Sn-S(n-1)=10^n-10^(n-1)=9*10^(n-1)n=1时,上式也成立