practice problems pending

Q1. Second principle of induction:
prove that for all natural numbers n (3 + sqrt(5))^n + (3 - sqrt(5))^n is an even integer.

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 is even.
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}is also even.
H.P.


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