by sum of a GP.
EDIT: Have I already posted this question? I have a feeling I already have.
I kinda rushed this, so hope I haven't made a mistake:
Alternatively:I kinda rushed this, so hope I haven't made a mistake:
Essentially if you continue, you end up removing this amount:
Which can be expressed as such:
Re-arranging:
Although you could arrive at this intuitively, a proper mathematical proof is probably better.
Here's a combinatorial proof for variety.
EDIT: Have I already posted this question? I have a feeling I already have.
that is a good question...not sure where to start lol
Yeah as you can see, it isn't necessarily hard, just a nice little result.I kinda rushed this, so hope I haven't made a mistake:
Essentially if you continue, you end up removing this amount:
Which can be expressed as such:
Re-arranging:
Although you could arrive at this intuitively, a proper mathematical proof is probably better.
Very nice. (the only reason why I ask for a possible combinatorial proof is because I want someone to show me one)Here's a combinatorial proof for variety.
Consider picking k+1 numbers out of the numbers {1,2,...,n+1}.
The right hand side of your equation is clearly equal to the number of ways of doing this.
Now for any given choice of k+1 numbers, the highest number chosen must be some j with k+1 =< j =< n+1. In each of these cases, we must select the remaining k numbers to be chosen from the j-1 numbers smaller than j.
Summing up the number of ways of doing this for j=k+1,...,n+1 yields the LHS of your equation.
As you have currently phrased the question, X and Y are both the point M itself...Let M be the midpoint of chord PQ in a circle
AB and CD are two other chords which pass through M
AB and CD intersect the chord PQ at points X and Y respectively
Show that M is the midpoint of XY
thanks for that , don't know what I was thinkingAs you have currently phrased the question, X and Y are both the point M itself...
It doesn't, as AB and CD are arbitrary the claim is false.thanks for that , don't know what I was thinking
hopefully it makes some sense now
Could you check this?It doesn't, as AB and CD are arbitrary the claim is false.
this IS 3u, I remember doing this in 3u, too easy for 4u m8Use mathematical induction and the product rule for differentiation to prove that
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx}(x^n)=nx^{n-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx}(x^n)=nx^{n-1}" title="\frac{d}{dx}(x^n)=nx^{n-1}" /></a>
For all positive integers n.
I believe this is more 3u than 4u, anyone else agree?