1.已知数列{an},{bn}满足a1=2,a2=4,bn=a(n+1)-an,b(n+1)=2bn+2.(1)求证:数列{bn+2}是公比为2的等比数列;(2)求an.注:a(n+1)中(n+1)为a的角标,后同.2.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12(
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![1.已知数列{an},{bn}满足a1=2,a2=4,bn=a(n+1)-an,b(n+1)=2bn+2.(1)求证:数列{bn+2}是公比为2的等比数列;(2)求an.注:a(n+1)中(n+1)为a的角标,后同.2.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12(](/uploads/image/z/1270305-9-5.jpg?t=1.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2C%7Bbn%7D%E6%BB%A1%E8%B6%B3a1%3D2%2Ca2%3D4%2Cbn%3Da%EF%BC%88n%2B1%EF%BC%89-an%2Cb%EF%BC%88n%2B1%EF%BC%89%3D2bn%2B2.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Bbn%2B2%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BA2%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82an.%E6%B3%A8%EF%BC%9Aa%EF%BC%88n%2B1%EF%BC%89%E4%B8%AD%EF%BC%88n%2B1%EF%BC%89%E4%B8%BAa%E7%9A%84%E8%A7%92%E6%A0%87%2C%E5%90%8E%E5%90%8C.2.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E4%B8%94a1%3D2%2Ca1%2Ba2%2Ba3%3D12%EF%BC%88)
1.已知数列{an},{bn}满足a1=2,a2=4,bn=a(n+1)-an,b(n+1)=2bn+2.(1)求证:数列{bn+2}是公比为2的等比数列;(2)求an.注:a(n+1)中(n+1)为a的角标,后同.2.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12(
1.已知数列{an},{bn}满足a1=2,a2=4,bn=a(n+1)-an,b(n+1)=2bn+2.
(1)求证:数列{bn+2}是公比为2的等比数列;
(2)求an.
注:a(n+1)中(n+1)为a的角标,后同.
2.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12
(1)求数列{an}的通项公式:
(2)令bn=anx^n(x∈R)求数列{bn}前n项和的公式
注:anx^n为an与x的n次幂的乘积,n为a的角标,后同.
1.已知数列{an},{bn}满足a1=2,a2=4,bn=a(n+1)-an,b(n+1)=2bn+2.(1)求证:数列{bn+2}是公比为2的等比数列;(2)求an.注:a(n+1)中(n+1)为a的角标,后同.2.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12(
1.(1)∵b(n+1)=2bn+2
∴b(n+1)+2=2bn+4
∴[b(n+1)+2]/(bn+2)=(2bn+4)/bn+2=2
∴数列{bn+2}是公比为2的等比数列
(2)由第一题可知bn+2=2*2^(n-1)
=2^n
∴bn=2^n-2
所以2^n-2=a(n+1)-an
a1=2
a2-a1=2
a3-a2=2^2
a4-a3=2^3
…………
an-a(n-1)=2^(n-1)
以上式子相加
得:an=2+2+2^2+2^3+……+2^(n-1)
=2+{2*[1-2^(n-1)]/(1-2)}
=2^n
2.(1)∵a1=2,a1+a2+a3=12
∴a1+a1+d+a1+2d=12
∴d=2
∴an=2+2(n-1)
2n
(2)sn=2*x+4*x^2+6*x^3+8*x^4+……2n*x^n
x*sn= 2*x^2+4*x^3+6*x^4+……(2n-2)*x^n+2n*x^(n+1)
两式相减
得(1-x)*sn=2x+2x^2+2x^3+2x^4+……2x^n-2n*x^(n+1)
={[2x(1-x^n)]/1-x}-2n*x^(n+1)
所以sn=……
你会算的吧