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There is an Arithmetic Progression in such way: $a_1=1$, and for each $n\in N$, $a_{n+1}=a_n+ \frac{1}{n(n+1)}$.

The problem: Prove in induction that for each $n\in N$, there exists: $a_n=2-\frac{1}{n}$. The steps of solving:

Prove that it's true for $n=1$:

$a_1=2-\frac{1}{1}=1$ True.

Suppose it's true for $k\in N$:

$a_k=2-\frac{1}{k}$ True.

Prove it's true for $k+1, k\in N$:

That's where I am stuck right now. Basically, It should be something like so: $a_{k+1}=2-\frac{1}{k+1}$

ancient mathematician
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Max Ilyouchenko
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1 Answers1

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$a_{k+1}=\color{blue}{a_k}+\frac{1}{k(k+1)}=\color{blue}{(2-\frac{1}{k})}+(\frac{1}{k}-\frac{1}{k+1})=2-\frac{1}{k+1}$

  • You proved $a_{n+1}=a_n+\frac{1}{n(n+1)}$, While what I need to prove is that $a_n = 2-\frac{1}{n}$ or would the induction answer be the same? – Max Ilyouchenko Sep 12 '20 at 15:44
  • I proved that if $a_k=2-\frac{1}{k}$ holds, then $a_{k+1}=2-\frac{1}{k+1}$ will also hold. That is the spirit of induction. –  Sep 12 '20 at 15:46
  • Wait, One question, How does this prove that it is true for each $n \in N$? – Max Ilyouchenko Sep 12 '20 at 15:53
  • You have already known that this is true for $n=1$, and we have proved that if it's true for $n=k$, it will also be true for $n=k+1$. So $n=2$ will do, $n=3$ will do, $n=4$ will do ..., etc. –  Sep 12 '20 at 16:39