早教吧 育儿知识 作业答案 考试题库 百科 知识分享

已知复数zn=an+bn•i,其中an∈R,bn∈R,n∈N*,i是虚数单位,且zn+1=2zn+.zn+2i,z1=1+i.(1)求数列{an},{bn}的通项公式;(2)求和:①a1a2+a2a3+…+anan+1;②b1b2-b2b3+b3b4-b4b5+…+(-1)n+1bnbn+1.

题目详情
已知复数zn=an+bn•i,其中an∈R,bn∈R,n∈N*,i是虚数单位,且zn+1=2zn+
.
zn
+2i,z1=1+i.
(1)求数列{an},{bn}的通项公式;
(2)求和:①a1a2+a2a3+…+anan+1;②b1b2-b2b3+b3b4-b4b5+…+(-1)n+1bnbn+1.
▼优质解答
答案和解析
(1)∵z1=a1+b1•i=1+i,∴a1=1,b1=1.
zn+1=2zn+
.
zn
+2i得an+1+bn+1•i=2(an+bn•i)+(an-bn•i)+2i=3an+(bn+2)•i,
an+1=3an
bn+1=bn+2
…(3分)
因此,数列{an}是以1为首项、公比为3的等比数列;数列{bn}是以1为首项、公差为2的等差数列,
可得,an=3n-1,bn=2n-1.…(6分)
(2)①由(1)知an=3n-1,∵
akak+1
ak-1ak
=32,
∴数列{anan+1}是以3为首项,公比为32的等比数列.
a1a2+a2a3+…+anan+1=
3(1-32n)
1-9
=
32n+1
8
-
3
8
.…(9分)
②当n=2k,k∈N*时,
b1b2-b2b3+b3b4-b4b5+…+(-1)n+1bnbn+1
=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2k-1b2k-b2kb2k+1
=-4b2-4b4-…-4b2k=-4(b2+b4+…+b2k
=-4•
k(b2+b2k)
2
=-8k2-4k=-2n2-2n
当n=2k+1,k∈N*时,
b1b2-b2b3+b3b4-b4b5+…+(-1)n+1bnbn+1
=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2k-1b2k-b2kb2k+1)+b2k+1b2k+2
=-8k2-4k+(4k+1)(4k+3)=2n2+2n-1

又∵n=1也满足上式,
b1b2-b2b3+b3b4-b4b5+…+(-1)n+1bnbn+1=
2n2+2n-1     当n为奇数时
-2n2-2n        当n为偶数时
…(14分)