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$$a_0=0$$

$$a_n=a_{n-1} + 2n^2-n$$

What I have so far, but I don't think it's right: $$x^n = x^{n-1} + 2x^2-x^{n-1}=2x+x^{n-2}-1$$

$$0=-x^{n-1}+x^{n-2}+2x-1$$

Michael Hardy
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Kristen M. Day
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  • @Kf-Sansoo The community asks that users show their work, so I'm not sure why you edited it out of the question. I've reverted it to show Kristen's thoughts. – Théophile Mar 27 '16 at 03:07
  • Your way of writing MathJax code is bizarre. I've attempted to clean it up, but I cannot be sure if the final result is what you intended. $\qquad$ – Michael Hardy Mar 27 '16 at 03:52
  • I'm still trying to figure out the best way to write the code, but the equations are what I had written down – Kristen M. Day Mar 27 '16 at 04:50

3 Answers3

2

hint: $a_n = (a_n-a_{n-1}) + (a_{n-1}-a_{n-2}) + \cdots + (a_1-a_0) + a_0 = \displaystyle \sum_{k=1}^n (2k^2-k)$. Can you find this sum?

DeepSea
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1

In this answer, it is shown that $$ \sum_{j=k}^n\binom{j}{k}=\binom{n+1}{k+1}\tag{1} $$ Since $a_n-a_{n-1}=2n^2-n$, we have the telescoping series $$ \begin{align} a_n &=a_0+\sum_{k=1}^n\left(a_k-a_{k-1}\right)\tag{2}\\ &=a_0+\sum_{k=1}^n\left(2k^2-k\right)\tag{3}\\ &=\sum_{k=1}^n\left[4\binom{k}{2}+\binom{k}{1}\right]\tag{4}\\[2pt] &=4\binom{n+1}{3}+\binom{n+1}{2}\tag{5}\\[5pt] &=\frac{(4n-1)(n+1)n}6\tag{6} \end{align} $$ Explanation:
$(2)$: Telescoping Series
$(3)$: apply the recurrence
$(4)$: write the summand using binomial coefficients
$(5)$: apply $(1)$
$(6)$: expand and factor

Community
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robjohn
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0

The homogenous part of the relation would be from $a_n=a_{n-1}$ with characteristic polynomial $(x-1)=0$, implying that the homogeneous solution would be of the form $\alpha\cdot 1^n$.

The non-homogenous part is a second degree polynomial (which is a special case of something of the form $1^n$). Ordinarily we would want to use an arbitrary second degree polynomial $\beta_2 n^2+\beta_1 n + \beta_0$ for the particular solution, however since the general solution is of the form $1^n$ and the particular solution is technically also of the form $\beta_2 n^21^n+\beta_1 n 1^n + \beta_0 1^n$, we must multiply by an additional $n$ to make this linearly independent from the other solution.

The solution will then be of the form:

$$a_n=\alpha \cdot 1^n + \beta_2 n^3 1^n + \beta_1 n^2 1^n + \beta_0 n 1^n = \alpha + \beta_0 n + \beta_1 n^2+\beta_2 n^3$$

I will leave solving for the coefficients of $\alpha, \beta_0, \beta_1, \beta_2$ to you to do, but this can be done using the recurrence relation and the initial conditions.

JMoravitz
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