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

Question

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:

$$\sum^{1}_{\color{blue}{k}=1}5\color{blue}{k}-4\rightarrow5\cdot\color{blue}{1}-4=1$$

$\color{red}{n}=1$, 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:

$$\sum^{p}_{k=1}5k-4=\frac12p(5p-3)$$

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

$$\sum^{p+1}_{k=1}5k-4=\frac12(p+1)(5(p+1)-3)$$

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

Thanks a lot in advance!

Answer 1

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

Now simplify to the right form

Answer 2

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

$\sum^{p+1}_{k=1}(5k-4)=\sum^{p}_{k=1}(5k-4)+5(p+1)-4=\frac12p(5p-3)+5p+1=\dfrac{p(5p-3)+10p+2}{2}$

That is

$\dfrac{5p^2-3p+10p+2}{2}=\dfrac{5p^2+7p+10}{2}=\dfrac{(p+1)(5p+2)}{2}=\dfrac{(p+1)(5(p+1)-3)}{2}$