A number theory problem by Akash Shukla

Number Theory Level pending

a n = a n 1 + a n + 1 a_{n}=a_{n-1}+a_{n+1} for n 1 n\geq1

a 0 = a 1 = 0 a_{0}=a_{1}=0

Find n = 1 a n = ? \sum_{n=1}^{\infty} a_{n} =?


The answer is 0.

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1 solution

Assume that a n = 0 a_{n} = 0 for some integer n.
a n = a n 1 a n 2 a_{n} = a_{n-1} - a_{n-2}
a n = 0 a n 1 = a n 2 a_{n} = 0 \to a_{n-1} = a_{n-2}
a n 1 = a n 2 = a n 2 a n 3 a n 3 = 0 a_{n-1}= a_{n-2} = a_{n-2} - a_{n-3} \to a_{n-3} = 0
Case 1:
n n is even.
a n = 0 a n 3 = 0 a n 6 = 0 a 1 = 0 a_{n} = 0 \to a_{n-3} = 0 \to a_{n-6} = 0 \ldots a_{1} =0 which is true.
Case 2:
n n is odd.
a n = 0 a n 3 = 0 a 0 = 0 a_{n} = 0 \to a_{n-3} = 0 \ldots \to a_{0} = 0 which is true as well.
a n = 0 , n Z + \therefore a_{n} = 0, n \in Z^{+}
n = 1 a n = 0 \therefore \displaystyle \sum_{n=1}^{\infty} a_{n} = 0


Nice one.I did it by taking values.

Akash Shukla - 4 years, 10 months ago

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