Let a 0 = 3, b 0 = 4, and c 0 = 5. Let a k = a k-1 + 2, b k = 2a k-1 + b k-1 + 2
ID: 2944772 • Letter: L
Question
Let a0 = 3, b0 = 4, and c0 = 5.
Let ak = ak-1 + 2, bk = 2ak-1 + bk-1 + 2, and ck = 2ak-1 + ck-1 + 2.
Prove that the numbers defined satisfy
ak2 + bk2 = ck2 for all natural numbers k.
There was a part of a of this question where I proved that ck - bk was constant for all k, but I don’t know if I need this or not.
What I have: Proof is by induction.
Show true for k=1. a12 + b12 = (a0 +2)2 + (2a0 +b0 + 2)2 = 132 = c12
Assume true for k=n: an2 + bn2 = cn2
Show true for k= n+1. This is where I get confused.
Please help.
Thank you
Explanation / Answer
ck - bk
= 2ak-1 + ck-1 + 2. - 2ak-1 - bk-1 - 2,
= ck-1 - bk-1
going like that.
ck-1 - b k-1 = ck-2 - bk-2 = 1.................
c0 - b0 = 1
= ck-1 - b k-1 =5 -4 = 1 which is constant.
ck= 1+bk
consider
an+1 ^2 + bn+1 ^2
= (ak + 2 )^2 + ( 2ak + bk + 2)^2
= ak^2 + 4ak + 4 + 4ak^2 + bk^2 + 4 + 8ak + 4bk + 4 akbk
= 4ak^2 + ak^2 + bk^2 + 4ak + 4 + 4 + 8ak + 4bk + 4 akbk
= 4ak^2 +ck^2 + 4ak + 4 + 4 + 8ak + 4bk + 4 akbk
= 4ak^2+ck^2 + 4 + 8ak + 4ak + 4 + 4bk + 4 akbk
ck+1 ^2 = (2ak + ck + 2.)^2
= 4ak^2 + ck^2+ 4 + 8ak + 4ck + 4akck
= 4ak^2 + ck^2+ 4 + 8ak + (4+ 4ak)ck
substitute ck = 1+bk
= 4ak^2 + ck^2+ 4 + 8ak + (4+ 4ak)(1+bk)
= 4ak^2 + ck^2+ 4 + 8ak + +4 + 4bk + 4ak + 4akbk
which is same as
= 4ak^2+ck^2 + 4 + 8ak + 4ak + 4 + 4bk + 4 akbk
= an+1 ^2 + bn+1 ^2
thus according to mathematical induction
given assumption is true.
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