My Infinite Sequence
Suppose there exists a sequence such that the nth term of the sequence is equivalent to the sum of all terms of the series, including that term itself. What is the 2013th term of the sequence?
The answer is 0.
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1 solution
You meant:
a 1 = a 1
a 2 = a 1 + a 2 ⇒ a 1 = 0 .
a 3 = 0 + a 2 + a 3 ⇒ a 2 = 0 .
a n = a n − 1 + a n ⇒ a n − 1 = 0 .
Thus, the sequence is a n = 0 , for every n ∈ N .
So a 2 0 1 3 = 0 .
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Yes. Sorry, I was on my iPad, so the formatting wasn't that good.
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No problem, dude :)
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Just a note, this problem trolled one of the top math students in San Diego. :)
u means first term is =to4
By definition, the first term = the first term(works for all numbers). Then, the second term = the second term + the first term(only works if the first term is equal to 0). The third term is equal to the third term + the second term + the first term. By substitution, the third term is equal to the third term plus the second term plus 0. From that, you can see that the second term is 0. Upon further investigation, you can find that the sequence is just 0, 0, 0, 0, ... Therefore, the 2013th term of the sequence is \boxed{0}.