设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式.

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设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式.

设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式.
设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3
若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式.

设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式.
因为an+1-an为等差数列,a2-a1=-2,a3-a2=-1解得公差为1,an+1-an=-2+(n-1)*1=n-3然后根据叠加法算an
a2-a1=-2,a3-a2=-1,.an-an-1=n-4,吧这些等式全部加在一起的an-a1=(-2+n-4)/2*n
利用同种方法先解bn+1-bn的通向公式再根据叠加法算bn
b2-b1=-2,b3-b2=-1,解得公比为1/2,所以bn+1-bn=-2*(1/2)^(n-1)
左边之和为bn-b1=-2*(1-(1/2)^n)/(1-1/2)(等比数列求和公式）

自己算

设{an+1-an}为{Tn},则T1=-2，T2=-1，d=1.首项为-2，即Tn=-(n+1)
同理得{bn+1-bn}=-2(-1)^(n-1)

你的an+1和bn+1里的1在外面还是里面啊？

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