数列{an}满足a1=0,a2=2,a(n+1)+a(n-1)=2(an+1),(n>=2),令bn=a(n+1)-an,1)求证{bn}为等差数列.2)求{an}的通项公式.
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数列{an}满足a1=0,a2=2,a(n+1)+a(n-1)=2(an+1),(n>=2),令bn=a(n+1)-an,1)求证{bn}为等差数列.2)求{an}的通项公式.
数列{an}满足a1=0,a2=2,a(n+1)+a(n-1)=2(an+1),(n>=2),令bn=a(n+1)-an,
1)求证{bn}为等差数列.
2)求{an}的通项公式.
数列{an}满足a1=0,a2=2,a(n+1)+a(n-1)=2(an+1),(n>=2),令bn=a(n+1)-an,1)求证{bn}为等差数列.2)求{an}的通项公式.
a(n+1)-an=an-a(n-1)+2
bn-b(n-1)=2
b1=a2-a1=2
bn=2n
a(n+1)-an=2n
an-a(n-1)=2(n-1)
.
a2-a1=2
an-a1=2(1+...+(n-1))
an=n(n-1)
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