Proof by mathematical induction $4+9+14+\ldots+(5n-1)=\frac{n(n+1)(2n+1)}2$

Question

I have to prove by mathematical induction that $L = R$

Question:

What I've tried so far:

I am completely stuck and don't know at which part I am failing. Can someone assist me?

Answer 1

You seem to be solving it like an equation. What you need to do is simply to show that:

$\displaystyle \frac{n(5n+3)} 2 + (5(n+1)-1)$

equals to:

$\displaystyle \frac{(n+1)(5(n+1)+3)}2$

by algebraically manipulating the first until you get to the final expression.

You cannot simply assume the final expression, but it helps a lot to write it down separately so you know what you're working toward.

First just expand and bring everything over a common denominator in the first expression:

$\displaystyle \frac{5n^2 + 13n + 8} 2$

You need to show you can make that into your final form.

Just factorise the quadratic:

$\displaystyle \frac{(n+1)(5n+8)} 2 $

and finally you have to "realise" you can rewrite $5n+8$ as $5(n+1)+3$ - this is motivated by "needing" to end up with a $+3$, which completes your process.

Answer 2

You made a mistake: when you wrote $n+1\bigl(5(n+1)+3\bigr)$, you should have written $(n+1)\bigl(5(n+1)+3\bigr)$. A proof can be done as follows:\begin{multline}\frac{n(5n+3)}2+5(n+1)-1=\frac{(n+1)\bigl(5(n+1)+3\bigr)}2\iff\\\iff5n^2+3n+10n+8=5n^2+13 n+8\iff\\\iff5n^2+13n+8=5n^2+13n+8.\end{multline}

Answer 3

We will proceed by mathematical induction. if $n=1$ then we have $5(1)-1=4$ and $\frac{1(5*1+3)}{2}=4$. Therefore, the proposition holds for $n=1$. Suppose that $ \sum_{j=1}^{k} (5j-1) = \frac{k(5k+3)}{2}$. Now, notice that $\sum_{j=1}^{k+1} (5j-1) =\sum_{j=1}^{k} (5j-1)+5(k+1)-1$ $$=\frac{k(5k+3)}{2}+5k+4 \\=\frac{5k^2+13k+8}{2}\\=\frac{(5k+8)(k+1)}{2}\\=\frac{(5(k+1)+3)(k+1)}{2}$$ as desired.