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Question

Given a1=12(a0+Aa0),a2=12(a1+Aa1)
and an+1=12(an+Aan)
Find n2 where a > 0, A > 0, prove that
anAan+(A)=(a1Aa1+(A))2n1
using mathematical induction.

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Solution

We have prove
anAan+(A)=(a1Aa1+(A))2n1 for all nϵN
Clearly for n = 1, equality (1) holds .
Now from the given relation a2=12(a1+Aa1)
we have
a2(A)=12(A)(a1+Aa1)=a21+A2a1(A)
Using componendo and dividendo , this gives
a2Aa2+(A)=a21+A2a1Aa21+A+2a1(A)=(a1Aa1+(A))2
Hence the equality (1) holds for n = 1,
Now assume that (1) holds for n = m(m2), that is, we assume
amAam+(A)=(a1Aa1+(A))2m1 ........(2)
Then by the given relation
an+1=12(an+Aan) we have
am+1(A)=12(A)(am+Aam) (m2) so that
am+1Aam+1(A)=12(am+Aam)A12(am+Aam)+A
=am22amA+Aam2+2amA+A
=(amAam+(A))2=(anAan+(A))2m by (2)
Hence the equality (1) holds for n = m + 1
Hence it holds for all nϵN

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