# Finding a closed form for $x_{k+1}=\frac{1}{3}+\frac{k-1}{2}+x_k$ [closed]

I have a sequence $\{x_n\}$ such that $x_2=\frac{7}{6}, x_3=\frac{5}{2}$ and $x_{k+1}=\frac{1}{3}+\frac{k-1}{2}+x_k$.

I want to find the $x_l$. I know that this is a problem of recurrence relation. But can not solve.

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## closed as off-topic by Thursday, Adriano, Hakim, le gâteau au fromage, GinaAug 2 at 0:22

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It appears that this sequence is very similar to $x_{k+1} = k + x_k$. What happens if you begin with that sequence? –  abiessu Jul 29 '13 at 5:04

## 2 Answers

Note that $$x_n-x_2=\sum_{k=3}^{n}(x_k-x_{k-1})=\frac{n-2}{3}+\frac{\sum_{k=3}^n(k-1)}{2}=\frac{n-2}{3}+\frac{\frac{n(n-1)}{2}-1}{2}$$ Now, you can find $x_n=\displaystyle \frac{n^2}{4}+\frac{n}{12}$ from this equation.

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$\frac{k^2}4+\frac{k}{12}$

Use the formula for the sum of the first $n$ integers, and note what adding $\frac{1}{3}$ $k$ times does.

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This gives the answer right away and does little to explain how it was obtained. Note the homework tag. –  dtldarek Jul 29 '13 at 5:08