The sequence a(n) is defined by: a(1)=20 ,a(2)=30, a(n+2)=3a(n+1) -a(n) ,n>=1 prove by induction that:?

Question

prove by induction that :
1+5a(n)*a(n+1) =[a(n) +a(n+1)]^2 +501

Answer 1

Assume you want to prove that for some statement P, P(n) is true for all n starting with n = 1. Induction requires just two steps:
(1) Prove that P(1) is true.
(2) Assuming that P(k) is true for some k. Derive from here that P(k+1) is also true.

First prove the statement is true for n=1.
1 + 5a(1)a(2) = [a(1) + a(2)]^2 + 501
a(1) = 20, a(2) = 30
1 + 5(20)(30) = [20 + 30]^2 + 501
1+ 3000 = 2500 + 501
3001 = 3001
So the statement is true for n = 1.

Now prove that the statement is true for n=n+1 assuming that it is true for n.

a(n + 2) = 3a(n + 1) - a(n)

1 + 5a(n)a(n + 1) = [a(n) + a(n + 1)]^2 + 501

LHS (used below for ease of writing)
RHS (used below for ease of writing)
LHS = RHS

Need to prove that if wer replace n by n+1 in the original statement that the resulting statement will be true.
1 + 5a(n+1)a(n + 2) = [a(n+1) + a(n + 2)]^2 + 501.... EQN-1

Left hand side EQN-1:
1 + 5a(n+1)a(n + 2) = 1 + 5a(n+1)[3a(n + 1) - a(n) ]
1 + 5a(n+1)a(n + 2) = 1 + 15[a(n+1)]^2 - 5a(n+1)a(n)
Add and subtract 10a(n+1)a(n) from the right hand side to get LHS from above.
1 + 5a(n+1)a(n + 2) = LHS + 15[a(n+1)]^2 - 10a(n+1)a(n)

New EQN-1.....Call this EQN-2
LHS + 15[a(n+1)]^2 - 10a(n+1)a(n) = [a(n+1) + a(n + 2)]^2 + 501
LHS = [a(n+1) + a(n + 2)]^2 + 501 - 15[a(n+1)]^2 + 10a(n+1)a(n)

Right hand side EQN-2:
[a(n+1) + a(n + 2)]^2 + 501 - 15[a(n+1)]^2 + 10a(n+1)a(n)
[a(n+1) + (3a(n + 1) - a(n))]^2 - 15[a(n+1)]^2 + 10a(n+1)a(n) +501

The first term:
[a(n+1)]^2 + 2a(n+1)(3a(n + 1) - a(n)) + [(3a(n + 1) - a(n))]^2
7[a(n+1)]^2 - 2a(n)a(n+1) + 9[a(n+1)]^2 - 6a(n)a(n+1) + [a(n)]^2
16[a(n+1)]^2 - 8a(n)a(n+1) + [a(n)]^2

All of the right hand side becomes with the above:
16[a(n+1)]^2 - 8a(n)a(n+1) + [a(n)]^2 - 15[a(n+1)]^2 + 10a(n+1)a(n) + 501
[a(n+1)]^2 + 2a(n)a(n+1) + [a(n)]^2 + 501
[a(n) + a(n+1)]^2 + 501 = RHS
So the right hand side of EQN-2 is RHS
And LHS = RHS

Answer 2

Base case:
n = 1
1 + 5a(1)*a(2) = [a(1) + a(2)]^2 + 501
1 + 5a(1)*a(2) = [a(1)^2 + 2a(1)a(2) + a(2)^2] + 501
1 + 5(20)(30) = 20^2 + 2(20)(30) + 30^2 + 501
1 + 3000 = 400 + 1200 + 900 + 501
3001 = 3001

Assume statement holds true for k such that:
1+5a(k)*a(k+1) =[a(k) +a(k+1)]^2 +501

Show statement holds for k+1 such that:
1+5a(k+1)*a(k+2) =[a(k+1) + a(k+2)]^2 +501
1+5a(k+1)*[3a(k+1)-a(k)] = [a(k+1) + 3a(k+1) - a(k)]^2 + 501
1+5a(k+1)*[3a(k+1)-a(k)] = [4a(k+1) - a(k)]^2 + 501
1+15a(k+1)^2 - 5a(k+1)a(k)] = [16a(k+1)^2 - 8a(k)a(k+1) + a(k)^2] + 501

Add and subtract 5a(k)a(k+1) to the LHS:
[5a(k)a(k+1) - 5a(k)a(k+1)] + 1+15a(k+1)^2 - 5a(k+1)a(k)] = [16a(k+1)^2 - 8a(k)a(k+1) + a(k)^2] + 501

5a(k)a(k+1) + 1+15a(k+1)^2 - 10a(k+1)a(k)] = [16a(k+1)^2 - 8a(k)a(k+1) + a(k)^2] + 501

Subtract (15a(k+1)^2 - 10a(k+1)a(k)) from both sides:
5a(k)a(k+1) + 1 = a(k+1)^2 + 2a(k)a(k+1) + a(k)^2 + 501

5a(k)a(k+1) + 1 = [a(k+1) + a(k)]^2 + 501
Which, by the hypothesis, is true.

So I have proven, by mathematical induction, that 1+5a(n)*a(n+1) =[a(n) +a(n+1)]^2 +501 for n ≥ 1

Answer 3

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