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Prove the following by mathematical induction

Let {${s_{n}}$} be the sequence defined by $s_{0}=\frac{\pi}{4}$ and $\forall{n}\geq1$, $s_{n}=s_{n-1}+\pi$.

Show: $\forall{n}\geq0$, $s_{n}=\frac{4n+1}{4}\pi$.

I think I should first show that $\forall{n}\geq1$, $s_{n}=s_{n-1}+\pi$ is just the same as $\forall{n}\geq0$, $\frac{4n+1}{4}\pi$, before I can proceed in proving $\forall{n}\geq0$,$s_{n}=\frac{4n+1}{4}\pi$ by mathematical induction. However, I had a hard time in showing that the two are just equivalent and with that, I can't start proving the latter.

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    $\begingroup$ $\forall{n}\geq0$, $\frac{4n+1}{4}\pi$ is not a complete mathematical sentence, like "Every day the temperature" is not a complete English sentence. $\endgroup$
    – Lee Mosher
    Sep 18, 2019 at 17:48

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Base case: $s_0=\frac{\pi}{4}=\frac{4\times0+1}{4}\pi$.

Induction step: Suppose the statement is true for $n$. Then $$s_{n+1}=s_n+\pi=\frac{4n+1}{4}\pi+\pi=\frac{4n+5}{4}\pi=\frac{4(n+1)+1}{4}\pi.$$

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  • $\begingroup$ The value of $s_0$ is already the base case. $\endgroup$
    – J.G.
    Sep 18, 2019 at 17:57

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