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The sum of the first $n$ terms of the series is $(4n + 5)^2$. Find the $n$th term of the series.

So far I have got $n_2=88, n_3=120, n_4=152, n_5=184 ...$

How should I continue?

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up vote 5 down vote accepted

Hint: If $S_n$ is the sum of the first $n$ terms and $t_n$ is the $n$th term, then observe that: $$ S_{n} - S_{n-1} = t_n $$

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Sorry, can you give me another hint please? I still don't get it. – Gez Bishop Aug 16 '13 at 0:24
Whoops made a typo. Anyways, what my hint is trying to say is: "To find the sum of the first $n$ terms, all I need to do is find the sum of the first $(n-1)$ terms, then add the $n$th term." Note that to find $S_{n-1}$, simply replace $n$ with $n-1$ to obtain: $$ S_{n-1} = (4(n-1)+5)^2 $$ – Adriano Aug 16 '13 at 0:28

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