Preditor.89.N7G said:
Ok here are some questions from the textbook that i dont queit understand.
Find
a) The co-efficent of x^3 in the expansion of (x^2-2/x)^3
b) The term independent of x in the expansion of (2x+1/x)^10
c) The co-efficent of x in the expansion of (2x^2-x^-1)^11
Use mathematical induction to prove
a) 1+2+3+4+5......+n= n(n+1)/2
b) 1x2x3x4x5.......nx(n+1)= (1/3)n(n+1)(n+2)
c) (19^2n)-1 is a multiple of 360
Thanks, i dont really understand the binomal theory very well, i'm a little bit confused. But i understand pascals triangle well, i know the binomal theory is in realtion to this but i dont know how exactly. Could someone explain easily how this is so....
Thanks again
I'm not very good at combinatorics, but I'm gonna give it a shot anyway
a) To get x^3, you need to multiply two terms of x^2 with one term of -2/x, and when you multiply the three brackets (of x^2-2/x), there are
3C
2=3 ways to get a term of x^3. And since x^2 * x^2 * (-2/x) gives x^3 with a coefficient of -2, the coefficent of x^3 in the expanded form will be -2*3 = -6.
b) Do this in a similar way to a). So to get a term independent of x, you need to multiple a term of 2x from one bracket to 1/x from another bracket, which gives 2. There are
10C
2 ways to do this, so the final coefficient is 2*
10C
2.
c) Try this before you look
For the induction ones, I'll just do the inductive step, you can prove the base case and write that blurb afterwards
a) If 1+...+k = k(k+1)/2,
1+...+k+(k+1) = k(k+1)/2 + (k+1) = [k(k+1)+2(k+1)]/2 = (k+1)(k+2)/2.
b) Uhh, this is not right...Take 5 for example, 5! = 120 =/= 4*5*6/3, nor does it equal to 5*6*7/3.
c) If 360|19
2k-1,
19
2(k+1)-1 = 19
2*19
2k-1 = 19
2*(19
2k-1) + 19
2 -1,
And since 360|19
2k-1, 360|19
2*(19
2k-1),
also 19
2 -1 = 360, 360|360,
Therefore 360|19
2(k+1)-1.
As for the interpretation of binomial theorem, think of (a+b)^n as the multiplication of n brackets of (a+b), then look at the thinking above for those combinatorics questions. So when you need a term of a
kb
n-k, you need to multiply terms of a, one each from k (out of the total n) brackets, to terms of b, one each from the remaining n-k brackets. Since there are
nC
k ways to choose which brackets to get the terms of a from (thus you get the terms of b from the remaining ones), and the coefficient of multiplying a's to b's is just 1 (so you get a term of a
kb
n-k for each way of multiplication), the coefficient of the term a
kb
n-k in the final expanded form is
nC
k.
Hopefully that made sense
EDIT: Oh, btw, if you want the connection to Pascal's triangle, see this:
n+1C
k+1 =
nC
k +
nC
k+1.
So if you know this, you can make sense of why kth number in nth row of Pascal's triangle is
nC
k, right? Then read what I wrote in the last paragraph above.
Now to prove
n+1C
k+1 =
nC
k +
nC
k+1:
Consider that you've got n+1 items, from which you need to choose k+1 items. Label the last item as L, so you could either include L in your choice, or exclude it.
If you include it, you need to choose another k items from the remaining n items, thus there are
nC
k ways of choice;
If you exclude it, you need to choose k+1 items from the remaining n items, thus there are
nC
k+1 ways of choice.
As either of these two are possible, you add them, giving
nC
k +
nC
k+1. This is equivalent to choosing k+1 items from all n+1 items, therefore
n+1C
k+1 =
nC
k +
nC
k+1.
