+1124 Hi friends!,
it's been a looooong time...
Please help me with the following:
T1
\({2 \over5}={1 \over3}+{1 \over15}\)
T2
\({2 \over7}={1 \over4}+{1 \over28}\)
T3
\({2 \over9}={1 \over5}+{1 \over45}\)
The want the formula for the nth fraction identity. I suppose one will just omit everything on the right of the "=" sign?
+1124 Hi Alan,
Thank tyou kindly...I have to admit, it's going to take me a while to understand it all....Thanx.
+118687 Practice will make you much better at these.
| n | 1 | 2 | 3 |
| Tn | \(\frac{2}{5}\) | \(\frac{2}{7}\) | \(\frac{2}{9}\) |
notice the pattern 5,7,9
n is only going up by 1 but the denominator of Tn is going up by 2
So a part of the denominor will be 2n
2n+3 works and the top is just 2. So \(T_n=\frac{2}{2n+3} \)
Now lets look at the next bit
\(T_n=\frac{1}{n+2}+\frac{1}{(n+2)(2n+3)}\\ \text{or if you prefer} \\ T_n=\frac{1}{n+2}+\frac{1}{2n^2+7n+6}\\\)
so
\(T_n=\frac{2}{2n+3} =\frac{1}{n+2}+\frac{1}{(n+2)(2n+3)}\\\)
+1124 Hi Melody!!..,
Thank you for explaining it, I will have to sit with this a while though to fully understand this. Thank you kindly..
+1124 Melody...no, I do not understand.....any of it....maybe I can comprehend the Tn value, but honestly, I cannot figure out how and WHY you got the rest..If you don't mind...would you kindly explain?
+118687 This is a really confusing question to start your learning process with.It is probably easier to think of it as 3 seperate questions.
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{2}{5}\) | \(\frac{2}{7}\) | \(\frac{2}{9}\) | |
| Tn | \(\frac{2}{2*\color{red}{1}\color{black}+3}\) | \(\frac{2}{2*\color{red}{2}\color{black}+3}\) | \(\frac{2}{2*\color{red}{3}\color{black}+3}\) | \(\frac{2}{2*\color{red}{n}\color{black}+3}\) |
I already tried to explain this bit. \(T_n=\frac{2}{2\color{red}{n}\color{black}+3}\)
Second one
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{1}{3}\) | \(\frac{1}{4}\) | \(\frac{1}{5}\) | |
| Tn | \(\frac{1}{\color{red}{1}\color{black}+2}\) | \(\frac{1}{\color{red}{2}\color{black}+2}\) | \(\frac{1}{\color{red}{3}\color{black}+2}\) | \(\frac{1}{\color{red}{n}\color{black}+2}\) |
Third one
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{1}{15}\) | \(\frac{1}{28}\) | \(\frac{1}{45}\) | |
| Tn | \(\frac{1}{5*3}\) | \(\frac{1}{7*4}\) | \(\frac{1}{9*5}\) | |
| Tn |
I already worked out how to get the those numbers from the first and second ones.
See if you can makes sense of what I have added here.
Like I said, you probably need to do a whole lot of easier ones to get the idea.
+1124 oh my goodness.....I would never have gotten it....How do you and Alan and the rest just KNOW all the answers?????????...it's completely beyond me.....
Thank you for making it clear...I will definitely practice that quite a bit before trying to explain to my pupil. God bless!!
+1124 Hi Melody,
looking at those you gave, the y value will be y+2
and f(n) will be f(n+5).....right?
+118687 Huh, you talking about the first one or the second one ?
Ill walk you through the first one
Sorry about the unblock sign I took a picture.
The x values are getting begger by 1
The y values are getting bigger by 2
So there will be 2x as a part of the answer.
I'll added another row inbetween to show you what I have so far
No to get from the 2x row to the y row I will have to subtract 1
y = 2x-1
Note
If you graph it, it will a line with gradient of 2 and y intercept of -1 .
Every point on that line will make this equation true.
I remember how much problems this initially gave me but once you get the hang of them they are not that hard. :)
Now you try the second one. And we should do a whole heap more after that.
+1632 Jurie, even if you didn't get the Tn shenanigans, you could still have looked for a pattern in the equations.
Example:
For T1, you notice that it is 2/5, and for T2, the denominator increases by 2.
The right side of the equation has a number that has a denominator that multiples the left side of the equation's denominator to get the very end term.
2/5 = 1/3 + 1/15.
The 5 * 3 = 15.
This pattern will continue forever.
You could also generalize this with a variable just like how Alan and melody described it.
The best idea to work with these is to get more practice.
