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Given that $(U_n)$ a numerical sequence such that :

$U_0$=3

$U_{n+1}=\sqrt{(U_n)^2+8n+16}$

Show that $(U_n)$ is a arithmetic progression.

So I have to show that $U_{n+1}-U_n=r$ where $r$ is a constant.

I started with this :

$(U_{n+1})^2-U_n^2=8n+16$

So $U_{n+1}-U_n=\frac{8(n+2)}{U_{n+1}+U_n}$

Also , for $n=0$ we have

$U_1=\sqrt{9+16}=5$

And $U_0=3$

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  • $\begingroup$ Its always a real number ! please check your question again. $\endgroup$ – DeepSea Jan 15 '16 at 12:18
  • $\begingroup$ @Kf-Sansoo yes its the general definition of an arithmetic sequence , right ? $\endgroup$ – user233658 Jan 15 '16 at 12:19
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    $\begingroup$ $r$ is constant, you should change $r$ to $c$ and state that $c$ is a constant. $\endgroup$ – DeepSea Jan 15 '16 at 12:20
  • $\begingroup$ @G.Sassatelli yes i can show that its numerical calculating U_1 snd U_2 .... $\endgroup$ – user233658 Jan 15 '16 at 13:25
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hint: $U_0 = 3, U_1 = 5, U_2 = 7 \Rightarrow U_n = 2n+3\Rightarrow U_{n+1} = ? \Rightarrow U_{n+1}-U_n = ?$

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By definition, we have $U_{n + 1} = |U_n + 4| = U_n + 4$, since $U_0 = 3 > 0$. We clearly have $U_n = 4n + 3$. This is an arithmetic progression, with $d = U_{n + 1} - U_n = 4$.

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