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I am asked to prove by induction that $4 + 9 + 14 + ... + 5n - 1 = \frac{n}{2}(3 + 5 n)$

As I understand it, during the induction step, one should replace $k$ by $k + 1$.

However, in the solution, the induction step is as follows

Let $n = k$

$4 + 9 + 14 + ... + (5k - 1) = \frac{k}{2}(3 + 5 k)$

Assume $n = k + 1$

$4 + 9 + 14 + ... + (5k - 1) + (5(k + 1) - 1) = \frac{k+1}{2}(3 + 5 (k+1))$

I don't understand on the LHS where the $(5k - 1)$ is coming from. Shouldn't $k$ be replaced by $k+1$, so that it looks like so

$4 + 9 + 14 + ... + (5(k + 1) - 1) = \frac{k+1}{2}(3 + 5 (k+1))$

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Pete
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    Do you know the sum symbol $\sum$, it would be much clearier – stity May 31 '17 at 13:07
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    The induction step must be: assume it true for $k$ and prove that it holds for $k+1$. – Mauro ALLEGRANZA May 31 '17 at 13:08
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    "Shouldn't $k$ be replaced by $k+1$": look closer, this is precisely what was done. –  May 31 '17 at 13:08
  • @YvesDaoust well, yes it should be, that is what confuses me. If you replace $k$ by $k+1$ in the line $let n = k$, then it should be as i showed on the last line. However, the actual answer is presented in $Assume n = k+1$, where a mysterious $5k-1$ appears in which the $k$ has not been replaced by $k+1$. Hope this clears up my problem. – Pete May 31 '17 at 13:12
  • Thus you start from the identity with $k$: $\ldots +5(k+1)=\frac k 2 (3+5k)$ and write the LHS for the case with $n=k+1$. It is the LHS of the previous formula with the "new term" for $n=k+1$ added. Then you rewrite it as $\frac k 2 (3+5k) + (5(k+1)-1)$ and you have to "manipulate" it to get the RHS. – Mauro ALLEGRANZA May 31 '17 at 13:14

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You simply misunderstand the notation a little.

$$4 + 9 + 14 + ... + (5(k + 1) - 1)$$

and

$$4 + 9 + 14 + ... + (5k - 1) + (5(k + 1) - 1)$$

are equivalent.

The writer just wanted to stress the addition of a term.