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marin [14]
2 years ago
12

Solve the following recurrence relation:

%200" id="TexFormula1" title="A_{n}=a_{n-1}+n; a_{1} = 0" alt="A_{n}=a_{n-1}+n; a_{1} = 0" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
-Dominant- [34]2 years ago
4 0

By iteratively substituting, we have

a_n = a_{n-1} + n

a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)

a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)

and the pattern continues down to the first term a_1=0,

a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))

\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)

\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k

Recall the formulas

\displaystyle \sum_{n=1}^N 1 = N

\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2

It follows that

a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2

\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1

\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}

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