practice problems pending

 Q1. Second principle of induction:

prove that for all natural numbers n, (3 + sqrt(5))^n + (3 - sqrt(5))^n is divisible by 2^n.

Let S_n = (3 + sqrt(5))^n + (3 - sqrt(5))^n

a = 3 + rt(5)
b = 3 - rt(5)

a+b = 6
ab = 4
a,b are roots of
x^2 - 6x + 4 = 0 => x^2 = 6x - 4
Multiply both sides with x^(n-1) =>
x^(n+1) = 6x^(n) - 4x^(n-1)
replace x with a,b and add them together
S_{n+1} = 6S_{n} - 4S_{n-1}
For n=1,2 Sn = 6,28, i.e. divisible by 2^1,2^2
Using second principle of induction assume it's true till S_{k} including S_{k-1} and below.
Then
S_{k+1} = 6S_{k} - 4S_{k-1}.
So it's divisible by 2^(k+1)
H.P.

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