已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
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已知数列{an}满足an=an+1-lg2,且a1=1,则通项公式为?已知数列{an}满足an=an+1-lg2,且a1=1,则通项公式为?已知数列{an}满足an=an+1-lg2,且a1=1,则通
已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
a(n+1) = a(n) + lg2
a(n) = a(n-1) + lg2
a(n-1) = a(n-2) + lg2
...
a(2) = a(1) + lg2
a(n+1) - a(1) + ( a(n) + a(n-1) + ...+ a(2) + a(1) ) = ( a(n) + a(n-1) + ...+ a(2) + a(1) ) + n*lg2
a(n+1) - a(1) = n*lg2
a(n+1) = n*lg2 + a(1) = ((n+1) - 1)*lg2 + a(1)
a(n) = (n -1)*lg2 + a(1) = (n-1)*lg2 + 1
a(n+1)-an=lg2
故为等差数列。
∴an=n*lg2+1-lg2
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