Question: These recurrences should be simple to solve but I see a ton of different ways to do it, such as general solution, particular solution etc. We did not talk about these in class, just need to get the general form of it. If anyone could help and explain (simple) would be great. Thank you!

$a\_{n} = a\_{n-1} + 2n + 1 , a\_{0}=1, a\_{1}=4$

Answer: 
And here is a completely different way of doing it - I'll leave you to work out for yourself what it means ;-)

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A [related problem](https://math.stackexchange.com/questions/775956/solving-recurrence-equation/776012#776012). Another way to go is

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> $$ a\_{n+1}-a\_n=2n+1 \implies \sum\_{i=0}^{n-1}(a\_{i+1}-a\_i)=\sum\_{i=0}^{n-1}(2i+1)=\dots\,. $$
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Other [techniques](https://math.stackexchange.com/questions/205372/how-to-solve-this-recurrence-relation-f-n-3f-n-1-12-1n/205491#205491).

**Note:** You need the following identity

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> $$ \sum\_{i=1}^{n} i = \frac{n(n+1)}{2}. $$
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