The sequence satisfies the above relation for all non-negative integers and with .
If , determine .
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For the sake of convenience, I'm rewriting the notation of a n as f ( n ) .
Given: f ( m + n ) + f ( m − n ) = 2 1 ( f ( 2 m ) + f ( 2 n ) )
Let m + n = X , m − n = Y
∴ 2 m = X + Y , 2 n = X − Y
∴ f ( X ) + f ( Y ) = 2 1 ( f ( X + Y ) + f ( X − Y ) )
Subsituting X = Y = 0 :
2 f ( 0 ) = f ( 0 ) ⟹ f ( 0 ) = 0
Substituting X = Y = 1 :
2 f ( 1 ) = 2 1 ( f ( 2 ) + f ( 0 ) ) ⟹ f ( 2 ) = 4
Subsituting Y = 1 :
f ( X ) + f ( 1 ) = 2 1 ( f ( X + 1 ) − f ( X − 1 ) )
⟹ f ( X + 1 ) = 2 f ( X ) − f ( X − 1 ) + 2
Substituting X = 2 gives f ( 3 ) = 9
It looks likes the following could be true: f ( X ) = X 2 ∀ X ∈ N ∪ 0
And this can be proved easily via induction, skipping the complete formal proof and just demonstrating the inductive step:
f ( X + 1 ) = 2 f ( X ) − f ( X − 1 ) + 2
⟹ f ( X + 1 ) = 2 X 2 − ( X − 1 ) 2 + 2
⟹ f ( X + 1 ) = 2 X 2 − X 2 + 2 X − 1 + 2 = X 2 + 2 X + 1 = ( X + 1 ) 2
∴ f ( X ) = X 2 ⟹ a 2 0 1 6 = f ( 2 0 1 6 ) = 2 0 1 6 2 = 4 0 6 4 2 5 6