已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
问题描述:
已知数列{an}满足an=an+1 -lg2,且a1=1,则通项公式为?
答
a(n+1) = a(n) + lg2a(n) = a(n-1) + lg2a(n-1) = a(n-2) + lg2...a(2) = a(1) + lg2a(n+1) - a(1) + ( a(n) + a(n-1) + ...+ a(2) + a(1) ) = ( a(n) + a(n-1) + ...+ a(2) + a(1) ) + n*lg2a(n+1) - a(1) = n*lg2a...