1. ## Need help simplifying/factoring

This problem is for a discrete math class, but requires factoring/simplifying. I haven't attempted anything like this in about 10 years.

I need $(\frac{(k(k+1))}{2})^2 + (k+1)^3$

to look like this: $(\frac{(k+1)(k+2)}{2})^2$

If that's possible. This may be asking alot, but a brief explanation of the simplification would be fantastic.

Thank you.

2. Originally Posted by webhound2
This problem is for a discrete math class, but requires factoring/simplifying. I haven't attempted anything like this in about 10 years.

I need $(\frac{(k(k+1))}{2})^2 + (k+1)^3$

to look like this: $(\frac{(k+1)(k+2)}{2})^2$

If that's possible. This may be asking alot, but a brief explanation of the simplification would be fantastic.

Thank you.
$(\frac{(k(k+1))}{2})^2 + (k+1)^3$

$= \frac{k^2(k+1)^2}{4} + \frac{(k+1)^3}{1}$

Now you can see that $(k+1)^2$ is a common factor, expression becomes:

$((k+1)^2)(\frac{k^2}{4} + \frac{k+1}{1})$

$=((k+1)^2)(\frac{k^2+4k+4}{4})$

$=(\frac{(k+1)(k+2)}{2})^2$

3. that's perfect. It seems so simple now. Thank you.

4. To give an explanation....

You want to get

$\displaystyle\left[\frac{k(k+1)}{2}\right]^2+(k+1)^3$

to

$\displaystyle\left[\frac{(k+1)(k+2)}{2}\right]^2$

$(k+1)^2$ is common and you want $2^2$ in the denominator

hence express $\displaystyle\ (k+1)^3=\frac{4(k+1)(k+1)^2}{4}=4(k+1)\left[\frac{k+1}{2}\right]^2$

Hence we get

$\displaystyle\ k^2\left[\frac{k+1}{2}\right]^2+4(k+1)\left[\frac{k+1}{2}\right]^2=\left(k^2+4k+4\right)\left[\frac{k+1}{2}\right]^2$

$=\displaystyle\ (k+2)^2\left[\frac{k+1}{2}\right]^2=\left[\frac{(k+1)(k+2)}{2}\right]^2$