设数列满足...a1=2 an 1=2a n 写出这个数列的前五项
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(Ⅰ)由已知,当n≥1时,an+1=[(an+1-an)+(an-an-1)+…+(a2-a1)]+a1=3(22n-1+22n-3+…+2)+2=22(n+1)-1.而a1=2,所以数列{an}的通
设bn=an/nSn=n^2-2n-2bn=sn-sn-1=2n-3b1=s1=-3所以an=n(2n-3)n>=2an=-3n=1
由于a1=-2,an+1=1−an1+an∴a2=1+a11−a1=−13,a3=1+a21−a2=12,a4=1+a31−a3=3,a5=1+a41−a4=−2=a1∴数列{an}以4为周期的数列∴
(1)根据题意,有An=(An-An-1)+(An-1-An-2)+…+(A2-A1)+A1=3-2^(2n-3)+3-2^(2n-5)+…+(3-2^3)+2再用分组求和法:=3n-【2^(2n-3
(1)∵a1+a22+a322+…+an2n-1=2n,n∈N*,①∴当n=1时,a1=2.当n≥2时,a1+a22+a322+…+an-12n-2=2(n-1),②①-②得,an2n-1=2.∴an
多写一项a1+2a2+2^2a3+...+2^n-2an-1=n-1/2,两式相减,有2^n-1an-2^n-2an-1=1/2,即2^nan-2^n-1an-1=1,所以2^nan=2a1+(n-1
令n=1时,a1=1*2*3=6;依题意:a1+2a2+3a3+.+nan=n(n+1)(n+2),a1+2a2+3a3+.+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)两式相减,得
a(n+1)-an=3*2^(2n-1)an-a(n-1)=3*2^(2n-3)...a3-a2=3*2^3a2-a1=3*2^1相加an-a1=3[2^1+2^3+2^5+2^7+...+2^(2n
由题意得:an-a(n-1)=3·2^(2n-3)a(n-1)-a(n-2)=3·2^(2n-5)..a2-a1=3·2^1叠加得:an-a1=3·[2^1+2^3+.+2^(2n-3)]注意:共n-
∵2nan+1=(n+1)an,∴a(n+1)/an=(n+1)/2n,∴a2/a1=2/2a3/a2=3/2×2a4/a3=4/2×3a5/a4=5/2×4……an/a(n-1)=n/2(n-1)两
(1)a1+3a2+…+3^(n-2)an-1=(n-1)/3a1+3a2+…+3^(n-1)an=(n-1)/3+3^(n-1)an=n/3an=(1/3)^n.(2)bn=n/an=n3^nSn=
1、①A1+3A2+3^2*A3+...+3^(n-1)*An=n/3,又A1+3A2+3^2*A3+...+3^(n-)*An-1=(n-1)/3,(比已知的式子最后少写一项,即有n-1项),两式相
a1+3a2+3²a3+…+3^(n-1)an=n/3a1+3a2+3²a3+…+3^(n-2)a(n-1)=(n-1)/3=n/3-1/3(n≥2)两式相减得:3^(n-1)an
前N项的和Sn加上第n+1项An+1,当然是前n+1项的和Sn+1咯
(1)证明:若an+1=an,即2an1+an=an,解得an=0或1.从而an=an-1=…a2=a1=0或1,与题设a1>0,a1≠1相矛盾,故an+1≠an成立.(2)由a1=12,得到a2=2
1、a(n+1)/an=(n+2)/(n+1)a(n+1)/(n+2)=an/(n+1)设cn=an/(n+1)则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1即c(n+1)=c
an满足an满足a1+2a2+3a3+...+nan=2^n所以有a1+2a2+3a3+...+(n-1)a(n-1)=2^(n-1)上面两式作减法有nan=2^n-2^(n-1)=2^(n-1)an
由递推式有a2-a1=3*2a3-a2=3*2*4a4-a3=3*2*4^2.an-a(n-1)=3*2*4^(n-2)累加得an-a1=2*4^(n-1)-8得an=2*4^(n-1)-6于是bn=
∵1=2,an+1=1+an1−an(n∈N*),∴a2=1+a11−a1=1+21−2=-3,a3=1+a21−a2=1−31+3=−12a4=1+a31−a3=1−121+12=13a5=1+a4
n=1时,3a1=3a1,n=2时,3+3a2=4a2,a2=33(a1+a2+a3+······+an)=(n+2)an①n>=2时有:3(a1+a2+a3+······+a(n-1))=(n+1)