My Infinite Sequence

Level 2

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

Tristan Shin
Dec 26, 2013

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}.

You meant:

a 1 = a 1 a_1 = a_1

a 2 = a 1 + a 2 a 1 = 0 a_2 = a_1 + a_2 \Rightarrow a_1 = 0 .

a 3 = 0 + a 2 + a 3 a 2 = 0 a_3 = 0 + a_2 + a_3 \Rightarrow a_2 = 0 .

a n = a n 1 + a n a n 1 = 0 a_n = a_{n-1} + a_n \Rightarrow a_{n-1} = 0 .

Thus, the sequence is a n = 0 a_n = 0 , for every n N n \in \mathbb{N} .

So a 2013 = 0 a_{2013} = 0 .

Guilherme Dela Corte - 7 years, 5 months ago

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Yes. Sorry, I was on my iPad, so the formatting wasn't that good.

Tristan Shin - 7 years, 5 months ago

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No problem, dude :)

Any problem you have on typing LaTeX, check out this rad website . It really will help a lot.

Guilherme Dela Corte - 7 years, 5 months ago

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@Guilherme Dela Corte Thanks!

Tristan Shin - 7 years, 5 months ago

Just a note, this problem trolled one of the top math students in San Diego. :)

Tristan Shin - 7 years, 2 months ago

u means first term is =to4

Shahida Khalid - 6 years, 7 months ago

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