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$a_{n+1}-a_n=2n+3, n \geq0; a_0=1$

First i solve the homogeneous equation:

$a_{n+1}-a_n=0$

I call the solution $a_{n}^p=\alpha, \alpha$ is a number.

Suppose $p_n=cn+d$ is a solution of $a_n$, then:

$$[c(n+1)+d]-[cn+d]=2n+3$$ $$c=2n+3$$

I'm stuck here, I'd appreciate your help.

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  • $\begingroup$ @Dr.MV Is not there an easier way? $\endgroup$ – retro_var Nov 14 '16 at 2:09
  • $\begingroup$ Try $p_n=cn^2+dn+e$. $\endgroup$ – Nicolas FRANCOIS Nov 14 '16 at 2:15
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Note that

$$\sum_{n=0}^{N-1}(a_{n+1}-a_n)=a_{N}-a_0 \tag1$$

and

$$\sum_{n=0}^{N-1}(2n+3)=N(N-1)+3N \tag2$$

Equating $(1)$ and $(2)$ reveals

$$a_n=(n+1)^2$$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Dec 2 '16 at 14:45

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