SOLUTION: Please explain this pdf step-by-step and preferably on the pdf. https://drive.google.com/file/d/1RDyhTvo7zrUekyy4UZ-fPXie1ygZIVVO/view?usp=drivesdk Thank you very much

Algebra ->  Rational-functions -> SOLUTION: Please explain this pdf step-by-step and preferably on the pdf. https://drive.google.com/file/d/1RDyhTvo7zrUekyy4UZ-fPXie1ygZIVVO/view?usp=drivesdk Thank you very much       Log On


   



Question 1203475: Please explain this pdf step-by-step and preferably on the pdf.
https://drive.google.com/file/d/1RDyhTvo7zrUekyy4UZ-fPXie1ygZIVVO/view?usp=drivesdk
Thank you very much

Found 4 solutions by greenestamps, Edwin McCravy, MathLover1, ikleyn:
Answer by greenestamps(13210) About Me  (Show Source):
You can put this solution on YOUR website!


We will not, on this forum, do your complete assignment for you; the rules of the forum say one problem per post. And I doubt that we have the capability of providing our answers on the link you provided.

I will explain the first problem for you. If you want help with the others, you need to post them individually -- entering each question using your keyboard instead of giving a link.

First problem then....

To understand binomial expansion, view the expression %28p%2Bq%29%5En as the product of n factors of p%2Bq.

In multiplying n factors of %28p%2Bq%29, each partial product is obtained by choosing either the p or the q from each of the n factors. That means that in each partial product the sum of the exponents on p and q has to be n.

To get a partial product that includes the expression p^n, you need to pick the p from each of the n factors.

To get a partial product that includes p^5, you need to pick the p from 5 of the n factors and the q from the other (n-5) factors.

Consider then a table of the term number, the exponents on p and q, and the coefficient of the term in the full expansion of %28p%2Bq%29%5En.
  term #  exponent on p   exponent on q  coefficient
 ----------------------------------------------------
    1          n               0          C(n,n) = C(n,0)    [you need to choose "p" in all n factors and "q" in none of them]
    2         n-1              1          C(n,n-1) = C(n,1)  [you need to choose "p" in (n-1) of the n factors and "q" in 1 of them]
    3         n-2              2          C(n,n-2) = C(n,2)  [you need to choose "p" in (n-2) of the n factors and "q" in 2 of them]
   ...
    n          1              n-1         C(n,1)  [you need to choose "p" in 1 of the n factors and "q" in (n-1) of them]
   n+1         0               n          C(n,0)  [you need to choose "p" in 0 of the n factors and "q" in all n of them]

Looking at the patterns in the table, we see that the k-th term contains expressions of

(1) q%5E%28k-1%29
(2) p%5E%28n-%28k-1%29%29 =p%5E%28n-k%2B1%29
(3) C%28n%2Ck-1%29

From that pattern we can see that the first term is p^n and the last term is q^n, so the first of the possible answer choices is true and the next three are not.

And from the pattern we can see that for the k-th = 4th term, k-1 is 4-1 = 3, so the 5th possible answer choice is true and the last one is not.


Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!

Here are the first 5 and the last 5 terms:

%28p%2Bq%29%5En%22%22=%22%22p%5En%22%22%2B%22%22np%5E%28n-1%29q%22%22%2B%22%22%28matrix%282%2C1%2Cn%2C2%29%29p%5E%28n-2%29q%5E2%22%22%2B%22%22%28matrix%282%2C1%2Cn%2C3%29%29p%5E%28n-3%29q%5E3%22%22%2B%22%22%28matrix%282%2C1%2Cn%2C4%29%29p%5E%28n-4%29q%5E4%22%22%2B%22%22%22%22%2A%22%22%2A%22%22%2A%22%22%22%22%2B%22%22%28matrix%282%2C1%2Cn%2Cn-4%29%29p%5E4%2Aq%5E%28n-4%29%22%22%2B%22%22%28matrix%282%2C1%2Cn%2Cn-3%29%29p%5E3%2Aq%5E%28n-3%29%22%22%2B%22%22%28matrix%282%2C1%2Cn%2Cn-2%29%29p%5E2q%5E%28n-2%29%22%22%2B%22%22%28matrix%282%2C1%2Cn%2Cn-1%29%29p%2Aq%5E%28n-1%29%22%22%2B%22%22q%5En

Edwin


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
binomial theorem
+ ... +++nC%28n-1%29%2A+x%5E1%2Ay%5E%28n-1%29+%2B+nCn+%2Ax%5E0%2Ay%5En

it is true that:
+1 st term is
nC0%2A+p%5En%2Aq%5E0+=1%2A+p%5En%2A1=p%5En+

the last term is
++nCn+%2Ap%5E0%2Aq%5En=q%5En+

+4 th term is
+nC3%2A+p%5E%28n-3%29%2A+q%5E3+


Answer by ikleyn(52897) About Me  (Show Source):
You can put this solution on YOUR website!
.

It is fantastically good business idea - in order for the volunteer tutors
make the answer cards for you, for free, immediately in pdf-format.

It looks like you are the first at this forum with such an advanced idea !

Hurry to patent the idea, before others get ahead of you !


Next time, as you get other similar brilliant business idea,
please report to us immediately !