# Finding $n^{th}$ term of sequence

$$3,8,17,32,57,\ldots$$ How do we find the $n^{th}$ term? I have been roaming aimlessly for a few minutes now. A definite pattern is in the differences of differences. Hints will be apreciated.

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$$3,8,17,32,57,\cdots$$

Taking the differences of consecutive terms $$5,9,15,25,\cdots$$

Taking the differences of consecutive terms $$4,6,10,\cdots$$

Taking the differences of consecutive terms $$2,4,\cdots,\text{ whose } n\text{ term } =2^n$$

The $n$ th term of $4,6,10,\cdots$ will be $2^n+(n$ th term of $2,2,2,\cdots)$ i.e., $2^n+2$

The $n$ th term of $5,9,15,25\cdots$ will be $2^n+2+(n$th term of $1,3,5,\cdots)$ i.e., $2^n+2+(2n-1)$

The $n$ th term of $3,8,17,32,57,\cdots$ will be $(2^n+2+2n-1)+(n$ th term of $-2,-1,2,7\cdots)$

Now the $n$ th term of $-2,-1,2,7\cdots$ can be found to be $(n-1)^2-2$ by similar difference method.

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Just so you know,there was a homework tag.But brilliant answer.Thanks for that. –  rah4927 Jun 9 '13 at 17:28
@rah4927, my pleasure.In spite of homework, I had to complete it as the most complexity lies in the few lines –  lab bhattacharjee Jun 9 '13 at 17:40
Yeah,I am not complaining though.I had figured out most of it before you posted it.But your answer provides completeness to the question. –  rah4927 Jun 9 '13 at 17:43
@rah4927, nice to hear that. My idea is to solve a problem in a generic way, so that we can solve most of the similar problems –  lab bhattacharjee Jun 9 '13 at 17:47
@rah4927, also this is a good link to recap artofproblemsolving.com/Wiki/index.php/… –  lab bhattacharjee Jun 9 '13 at 17:49
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From OEIS, the sequence is $$2^n + n^2$$

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I was hovering along the same direction almost.Thanks for the answer. –  rah4927 Jun 9 '13 at 16:55