7
$\begingroup$

I have a small algebra problem

If the sum of two consecutive integers is $x$, what would be their product?

I've made equation.$$n+(n+1)=x$$ But, don't understand how to find their sum.

$\endgroup$
1
  • 1
    $\begingroup$ So, $n+(n+1)=x$ reworded is $2n=x-1$ or $n=\frac{x-1}{2}$, adding one to each side we get another equation. So we have $n\cdot (n+1)=\dots$ $\endgroup$
    – JMoravitz
    Commented Mar 30, 2017 at 6:29

4 Answers 4

10
$\begingroup$

$$n + (n + 1) = x \implies n = \frac{x - 1}{2}$$

Then $$n(n + 1) = \left(\frac{x - 1}{2}\right) \left(\frac{x - 1}{2} + 1\right) = \left(\frac{x - 1}{2}\right)\left(\frac{x + 1}{2}\right) = \frac{x^2 - 1}{4}.$$

$\endgroup$
5
$\begingroup$

Now find the value of $n$ from this equation.$n+(n+1)=x\\2n+1=x\\n=\dfrac{x-1}{2}\tag*{}$Therefore,$n(n+1)=\dfrac{x-1}{2}\left(\dfrac{x-1}{2}+1\right)\\=\dfrac{x^2-1}{4}\tag*{}$Just as a quick check: the sum of $4$ and $5$ is $9$, and their product is $20$ which is indeed $(9^2-1)/4$.

$\endgroup$
5
$\begingroup$

You've "made" the equation $n + (n + 1) = x$ but don't understand how to find their sum? I think you mean you don't understand how to find the product $n(n + 1)$.

The equation you've "made" is an important clue. If $n$ is odd, then $n + 1$ is even, otherwise $n$ is even and $n + 1$ is odd. Either way, $x$ is odd. Therefore, $$n = \frac{x - 1}{2}, n + 1 = \frac{x - 1}{2} + 1.$$ If you know how to multiply polynomials, you should be able to take it from here.

$\endgroup$
1
$\begingroup$

HINT:

$$4n(n+1)=(n+1+n)^2-(n+1-n)^2$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .