Prove by induction that the following equation is true for $n\ge1$ $$\sum^{n}_{k=1}5k-4=\frac12n(5n-3)$$

I did the following:


$\color{red}{n}=1$, so: $$\frac12\cdot\color{red}{1}(5\cdot\color{red}{1}-3)=1$$

I have now proven that for $n=1$, the statement is true

I now assume that the formula is correct for $n=p$, so:


I now have to prove that the statement holds for $p+1$:


I don't have any idea how to continue...

Thanks a lot in advance!

  • 2
    $\begingroup$ $\sum^{p+1}_{k=1}5k= 5(p+1)+\sum^{p}_{k=1}5k$ $\endgroup$ – Surb Jan 6 '15 at 15:27
  • $\begingroup$ See Faulhaber's formulas. $\endgroup$ – Lucian Jan 6 '15 at 17:50

What you need to do is start with the $p+1$ statement and try to isolate the $p$ statement out of it.


That is



$$\sum_{k=1}^{p+1} (5k-4) = [5(p+1)-4] + \sum_{k=1}^{p} (5k-4)$$ $$\underbrace{=}_{I.H.} \ [5(p+1)-4] + \dfrac{1}{2}p(5p-3)$$

Now simplify to the right form


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