Let {an} be a sequence of positive numbers such that
a1 + a2 + ... + an = [(1+an)/2]^2 for n = 1,2,3...
Prove by induction that an = 2n - 1 for n = 1,2,3...
thank you for yr help~
pure maths ( M.I.)
2009-09-19 4:52 am
回答 (2)
2009-09-19 5:24 am
2009-09-19 5:25 am
for n=1,
a1 = [(1+a1)/2]^2
a1 = 1 = 2(1)-1 = 1
n=1 is true
asssume n=k is true, ak=2k-1
for n=k+1,
a1 + a2 + ... + ak + ak+1 = [(1+ak+1)/2]^2
[(1+ak)/2]^2 + ak+1 = [(1+ak+1)/2]^2
[(1+2k-1)/2]^2 + ak+1 = [(1+ak+1)/2]^2
ak+1 = 1+2k or 1-2k (rejected, as k = 1, 2, 3, ...)
ak+1 = 2(k+1)-1
n=k+1 is true
a1 = [(1+a1)/2]^2
a1 = 1 = 2(1)-1 = 1
n=1 is true
asssume n=k is true, ak=2k-1
for n=k+1,
a1 + a2 + ... + ak + ak+1 = [(1+ak+1)/2]^2
[(1+ak)/2]^2 + ak+1 = [(1+ak+1)/2]^2
[(1+2k-1)/2]^2 + ak+1 = [(1+ak+1)/2]^2
ak+1 = 1+2k or 1-2k (rejected, as k = 1, 2, 3, ...)
ak+1 = 2(k+1)-1
n=k+1 is true
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