Get this equation to equal another

[crybloodwing]

So for a proof, I have to get

(-5k)/(2k+1) + (5)/(1-4(k+1)^2) to equal (-5(k+1))/(2(k+1)+1)

I have tried ways such as distributing and multiplying to get a common denominator, then trying to simplify, but I am unable to get to the final equation.

 

[Steven G]

First of all an equation has an equal sign. So you are not trying to show that one equation equals another equation.
You said that you tried different things. That is wonderful! Now can we please see some of what you tried so that we can guide you to the correct answer

 

[crybloodwing]

First of all an equation has an equal sign. So you are not trying to show that one equation equals another equation.
You said that you tried different things. That is wonderful! Now can we please see some of what you tried so that we can guide you to the correct answer

Actually, for the induction proof, I do need to show that I get to an equation that equals/matches my equation for p(k+1)

I was able to get it when I used other variables under closure of integers, but I need the variable to be k, the original variable.

 

[Dr.Peterson]

I would do what you said to the first expression, combining it into a single fraction using a common denominator and then simplifying. Then I would do the same to the second expression. If the results are equal, then you have shown what you need to show.

Technically, you will have shown that the two expressions are equal on the intersection of their domains; the domains are not the same, so the expressions are not really equal. (Now that you have shown the context of your question, we can see that you only need to prove equality for positive integers, so this is okay.)

It looks like you failed to simplify (cancel a common factor), which complicated things. It should be obvious that you will need to simplify, since you know the denominator of the target expression.

 

[Steven G]

I would write (-5k)/(2k+1) + (5)/(1-4(k+1)^2) with a common denominator. Factor the denominator in the 2nd fraction
Let's see what you get.

 

[crybloodwing]

I would write (-5k)/(2k+1) + (5)/(1-4(k+1)^2) with a common denominator. Factor the denominator in the 2nd fraction
Let's see what you get.

It would be [-5k(2k+3) - 5]/(2k+1)(2k+3), which I have also worked with.

Got it. Thanks!

 

[MarkFL]

This is how I worked it...

It appears we begin with the expression:

[MATH]\frac{-5k}{2k+1}+\frac{5}{1-4(k+1)^2}[/MATH]
I would begin by factoring the second denominator as the difference of squares:

[MATH]\frac{-5k}{2k+1}+\frac{5}{(1+2(k+1))(1-2(k+1))}[/MATH]
Distribute:

[MATH]-\frac{5k}{2k+1}-\frac{5}{(2k+3)(2k+1)}[/MATH]
Factor out [MATH]-\frac{5}{2k+1}[/MATH]
[MATH]-\frac{5}{2k+1}\left(k+\frac{1}{2k+3}\right)[/MATH]
Combine terms with brackets:

[MATH]-\frac{5}{2k+1}\left(\frac{2k^2+3k+1}{2k+3}\right)[/MATH]
Factor numerator within brackets:

[MATH]-\frac{5}{2k+1}\left(\frac{(2k+1)(k+1)}{2k+3}\right)[/MATH]
Divide out common factors:

[MATH]-5\left(\frac{k+1}{2k+3}\right)[/MATH]
This is equivalent to:

[MATH]-\frac{5(k+1)}{2(k+1)+1}[/MATH]