Why Does the Binomial Theorem Solution Differ from the Book's Answer?

[gaganspidey]

Homework Statement

Homework Equations

Formula => C(n,r) or ⁿC_r =n!/r!(n-r)! & the basic Binomial Theorem formula.

*Answer mentioned in book = nx

The Attempt at a Solution

The LHS should be (x+y)ⁿ & the given question is its expansion only if that 'r' is not multiplied in the question. I also know how to add using the summation feature but I am not able to get x+y in the expansion (so I could substitute the value of 1). If I substitute 1 on the LHS I get => 1ⁿ.

 

[tiny-tim]

Welcome to PF!

Hi gaganspidey! Welcome to PF!

Answer mentioned in book = nx

hmm … when you see the trick, you're going to kick yourself about how simple it is.

Hint: the sum of n elements is x times n … think "average" …

(and everyone else, please no more hints until he's had a chance to reply! )​

 

[gaganspidey]

It might sound strange but I still don't get it. From your hint what I've been able to understand is that there are n elements in the expansion because r ranges from 0 to n, but what has x got to do with it, I mean isn't it something like this -

0 + nxy^n-1 +n(n-1)x²y^n-2 + ... + xⁿ

So how do we solve it to get the answer.

So sorry for being silly, I've a feeling its really easy but I am just not able to get it

 

[tiny-tim]

Sorry … perhaps that was a bit too cryptic

Try this: imagine you do the same thing n times (eg, try to hit a target), with independent results, and the probability of success each time is x:

what does ∑ rⁿC_rx^ry^n-r represent?

 

[gaganspidey]

So its means we don't actually have to calculate anything here, we just have to assume its average, like you said earlier,i.e, :-

= 0 + nxy^n-1 +n(n-1)x²y^n-2 + ... + xⁿ <-----(the first actual step)
= x + x + x + x + ... + x upto n terms <-----(assuming we get x by solving each term on average)
= xn

But then why have the given value 'x+y=1' not been used, I think it tells us that the values of 'x' & 'y' are less than 1. And why do we take the average of each term as 'x' and not as any other term such as 'y' or 'r'

 

[tiny-tim]

Try this: imagine you do the same thing n times (eg, try to hit a target), with independent results, and the probability of success each time is x:

what does ∑ rⁿC_rx^ry^n-r represent?

∑ rⁿC_rx^ry^n-r represents the expected value of the number of successes (after n independent tries).

But if you haven't done "expected values", try this:

What is ∑ (r/nx) ⁿC_rx^ry^n-r ?

(expand the ⁿC_r)

 

[gaganspidey]

Unfortunately I'm not aware of "expected value" concept.
Here's the expansion that I've done :
= 0 + (1/nx) n(n-1)!/(n-1)! xy^n-1 + (2/nx) n(n-1)(n-2)!/2!(n-2)! x²y^n-2 + ... + (n/nx) ⁿC_nxⁿy^n-n <------(substituting r=0,1,2...n)
= y^n-1 + (n-1)xy^n-2 + ... + x^n-1

 

[jambaugh]

I find that for general problems like this you need to manipulate the binomial coefficient expressed as a ratio of factorials(absorb the r term into it).

Note:
$$r! = r\cdot (r-1)! = r\cdot (r-1)\cdot (r-2)! ...$$

Thus:

$$\sum_{r=0}^{n} r \mathbf{C}^n_r x^r y^{n-r}=\sum_{0}^{n} r \cdot\frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$=\sum_{r=0}^{n} r \cdot\frac{n!}{r\cdot(r-1)!(n-r)!} x^r y^{n-r}$$

Cancel the r's and then manipulate until you get a different binomial coefficient. Then see if you can factor out x's or y's from the sum to get another different binomial expansion. Then pay close attention to the end terms on the summation. See if you can manipulate it into something plus something times another binomial expansion sum.

Note that you can recursively solve this with higher powers of r since:

$$\sum_{r=0}^{n} r^2 \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$\sum_{r=0}^{n} r(r-1) \frac{n!}{r!(n-r)!} x^r y^{n-r}-\sum_{0}^{n} r \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

[EDIT: I forgot to write r=0 instead of 0 in the summation notation. Fixed now.]

 

[tiny-tim]

0 + (1/nx) n(n-1)!/(n-1)! xy^n-1 + (2/nx) n(n-1)(n-2)!/2!(n-2)! x²y^n-2 + ...

hold it! what's going on here?

you're getting really confused …

n(n-1)(n-2)/2!(n-2)! isn't ⁿC₂ …

(I assume you meant that, without the extra "!")

you either mean n(n-1)/2!, or you mean n!/2!(n-2)!

Try again, multiplying by r/nx, and just using n!/r!(n-r)! (and without breaking it down any further)

 

[Дьявол]

It might sound strange but I still don't get it. From your hint what I've been able to understand is that there are n elements in the expansion because r ranges from 0 to n, but what has x got to do with it, I mean isn't it something like this -

0 + nxy^n-1 +n(n-1)x²y^n-2 + ... + xⁿ

So how do we solve it to get the answer.

So sorry for being silly, I've a feeling its really easy but I am just not able to get it

Ok, you're on the right track.

What don't you use the sum of geometric series?

$$a + ar + a r^2 + ar^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^n}{1-r},$$

Regards.

 

[mathie.girl]

I find that for general problems like this you need to manipulate the binomial coefficient expressed as a ratio of factorials(absorb the r term into it).

Note:
$$r! = r\cdot (r-1)! = r\cdot (r-1)\cdot (r-2)! ...$$

Thus:

$$\sum_{r=0}^{n} r \mathbf{C}^n_r x^r y^{n-r}=\sum_{0}^{n} r \cdot\frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$=\sum_{r=0}^{n} r \cdot\frac{n!}{r\cdot(r-1)!(n-r)!} x^r y^{n-r}$$

Cancel the r's and then manipulate until you get a different binomial coefficient. Then see if you can factor out x's or y's from the sum to get another different binomial expansion. Then pay close attention to the end terms on the summation. See if you can manipulate it into something plus something times another binomial expansion sum.

I think this is an excellent way of doing this! Since neither n nor x are involved in the summation index, you can factor both of them out. And what's left can be manipulated into something much nicer.

Note that you can recursively solve this with higher powers of r since:

$$\sum_{r=0}^{n} r^2 \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$\sum_{r=0}^{n} r(r-1) \frac{n!}{r!(n-r)!} x^r y^{n-r}-\sum_{0}^{n} r \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

[EDIT: I forgot to write r=0 instead of 0 in the summation notation. Fixed now.]

What do you mean by this? Apparently I'm a bit out of it and I'm not sure what you're suggesting. Thanks!

 

[Gib Z]

Alternatively, we could notice that the sum is the derivative of the usual binomial expansion multiplied by x.

 

[gaganspidey]

hold it! what's going on here?

you're getting really confused …

n(n-1)(n-2)/2!(n-2)! isn't ⁿC₂ …

(I assume you meant that, without the extra "!")

you either mean n(n-1)/2!, or you mean n!/2!(n-2)!

Try again, multiplying by r/nx, and just using n!/r!(n-r)! (and without breaking it down any further)

Yes actually I expanded it a little bit further, but when applying the formula only, I get :

= 0 + (1/nx) n!/(n-1)! xy^n-1 + (2/nx) n!/2!(n-2)! x²y^n-2 + ... + (n/nx) ⁿC_nxⁿy^n-n <------(substituting r=0,1,2...n)
= y^n-1 + (n-1)xy^n-2 + ... + x^n-1

which is the same result.

 

[gaganspidey]

Ok, you're on the right track.

What don't you use the sum of geometric series?

$$a + ar + a r^2 + ar^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^n}{1-r},$$

Regards.

I wonder how I could apply this formula because we need a geometric progression for it,ie, a₂/a₁=a₃/a₂ whereas I am not getting LHS = RHS.

 

[gaganspidey]

I find that for general problems like this you need to manipulate the binomial coefficient expressed as a ratio of factorials(absorb the r term into it).

Note:
$$r! = r\cdot (r-1)! = r\cdot (r-1)\cdot (r-2)! ...$$

Thus:

$$\sum_{r=0}^{n} r \mathbf{C}^n_r x^r y^{n-r}=\sum_{0}^{n} r \cdot\frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$=\sum_{r=0}^{n} r \cdot\frac{n!}{r\cdot(r-1)!(n-r)!} x^r y^{n-r}$$

Cancel the r's and then manipulate until you get a different binomial coefficient. Then see if you can factor out x's or y's from the sum to get another different binomial expansion. Then pay close attention to the end terms on the summation. See if you can manipulate it into something plus something times another binomial expansion sum.

Note that you can recursively solve this with higher powers of r since:

$$\sum_{r=0}^{n} r^2 \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$\sum_{r=0}^{n} r(r-1) \frac{n!}{r!(n-r)!} x^r y^{n-r}-\sum_{0}^{n} r \frac{n!}{r!(n-r)!} x^r y^{n-r}$$

[EDIT: I forgot to write r=0 instead of 0 in the summation notation. Fixed now.]

Hmm...got to give it a try.

 

[gaganspidey]

Alternatively, we could notice that the sum is the derivative of the usual binomial expansion multiplied by x.

Sorry, I still have to learn derivation/differentiation.

 

[tiny-tim]

Yes actually I expanded it a little bit further, but when applying the formula only, I get :

= 0 + (1/nx) n!/(n-1)! xy^n-1 + (2/nx) n!/2!(n-2)! x²y^n-2 + ... + (n/nx) ⁿC_nxⁿy^n-n <------(substituting r=0,1,2...n)
= y^n-1 + (n-1)xy^n-2 + ... + x^n-1

which is the same result.

Nooo …

you're trying to guess what it is from what the first couple of terms look like …

that's a recipe for disaster …

just deal with one (general) term: what is (r/nx)ⁿC_r ?

(this is the same as jambaugh's suggestion …)

$$\sum_{r=0}^{n} r \mathbf{C}^n_r x^r y^{n-r}=\sum_{0}^{n} r \cdot\frac{n!}{r!(n-r)!} x^r y^{n-r}$$

$$=\sum_{r=0}^{n} r \cdot\frac{n!}{r\cdot(r-1)!(n-r)!} x^r y^{n-r}$$

Cancel the r's and then manipulate until you get a different binomial coefficient. Then see if you can factor out x's or y's from the sum to get another different binomial expansion.

(and ignore Дьявол's suggestion)

 

[Gib Z]

Sorry, I still have to learn derivation/differentiation.

Ahh sorry, I forgot it was in the pre calc forums. Pity, I thought it was a pretty short and simple solution =[

EDIT: I guess it depends on what you can see from inspection. I mean, if you can see that your series is the expansion of $nx (x+y)^{n-1}$ then that would be great lol! But obviously it needs practice for these kind of things.

 

[jambaugh]

Hmm...got to give it a try.

I worked it out and there are a couple of "tricks" trying from the start (such as adding and subtracting 1 and grouping to get a particular form.)

I suggest you also give the binomial expansion of:
$$(x+y)^{n-1}$$
so you can work from both ends.

Also it is quite helpful to pick some small values of n say 3, 4 and 5 and working through theses examples explicitly to see what's going on in general. (Good technique in any such problem!)

 

[gaganspidey]

I worked it out and there are a couple of "tricks" trying from the start (such as adding and subtracting 1 and grouping to get a particular form.)

I suggest you also give the binomial expansion of:
$$(x+y)^{n-1}$$
so you can work from both ends.

Also it is quite helpful to pick some small values of n say 3, 4 and 5 and working through theses examples explicitly to see what's going on in general. (Good technique in any such problem!)

This went over my head

 

[tiny-tim]

What is (r/n)ⁿC_r ?

 

[gaganspidey]

What is (r/n)ⁿC_r ?

Do you mean : -

$$\sum_{r=0}^{n}$$ r/(nx) ⁿC_r

= $$\sum_{r=0}^{n}$$ r/(nx)n!/r(r-1)!(n-r)!

= $$\sum_{r=0}^{n}$$ 1/(nx)n!/(r-1)!(n-r)! <-----(cancelling the r's)

= 0 + 1/(nx) n!/(n-1)! + 1/(nx) n!/(n-2)! +...+ 1/(nx) n!/(n-1)! <----(putting r=0,1,2...n)

= 1/x + (n-1)/x + (n-1)(n-2)/2x + ... +1/x

 

[tiny-tim]

Do you mean : -

$$\sum_{r=0}^{n}$$ r/(nx) ⁿC_r

= $$\sum_{r=0}^{n}$$ r/(nx)n!/r(r-1)!(n-r)!

= $$\sum_{r=0}^{n}$$ 1/(nx)n!/(r-1)!(n-r)! <-----(cancelling the r's)

ok so far … but don't go beserk …

(and forget the ∑ … I didn't ask about the ∑)

now what else can you cancel?

and then what is the result equal to? ​

 

[jambaugh]

This went over my head

$$\sum_{r=0}^3 r\cdot C_r^3 \cdot x^r y^{3-r}$$
$$= 0\cdot 1\cdot x^0 y^3 + 1\cdot 3 \cdot x^1 y^2 + 2\cdot 3 \cdot x^2 y^1 + 3\cdot 1\cdot x^3$$
$$= 3 \cdot x y^2 + 6 \cdot x^2 y + 3\cdot x^3$$
$$= 3x( y^2 + 2 \cdot x y + \cdot x^2)$$
$$= 3x(x+y)^2 = 3x$$

Now try it with n=4.

[edit] Another helpful trick when dealing with the general n sum. Where you have n-r you can rewrite that as n-r + 1 - 1 and regroup the 1's...

 

[jambaugh]

Do you mean : -

$$\sum_{r=0}^{n}$$ r/(nx) ⁿC_r

=
...

I suggest you put the whole equations inside the tex:

$$\sum_{r=0}^{n} r/(nx) C^n_r$$

$$=\sum_{r=0}^{n} r/(nx)n!/r(r-1)!(n-r)!$$


And use the \frac{ numerator}{denominator} latex command so your fractions are clear.


$$\sum_{r=0}^{n} \frac{r}{nx} C^n_r$$

= $$\sum_{r=0}^{n} \frac{r}{nx} \frac{n!}{r(r-1)!(n-r)!}$$


You have forgotten about the x^r y^{n-r} factor. You need it.
$$=\sum_{r=0}^{n} \frac{r}{nx} \frac{n!}{r(r-1)!(n-r)!} x^r y^{n-r}$$
At this point you'll need a trick or two once you simplify.
Remember you can also write the n! as n(n-1)! and the n-r as n-1 - r + 1 = n-1 -(r-1).

But again I suggest you go through some sums with n=4 and n=5 to see why all these manipulations.

 

[gaganspidey]

$$\sum_{r=0}^3 r\cdot C_r^3 \cdot x^r y^{3-r}$$
$$= 0\cdot 1\cdot x^0 y^3 + 1\cdot 3 \cdot x^1 y^2 + 2\cdot 3 \cdot x^2 y^1 + 3\cdot 1\cdot x^3$$
$$= 3 \cdot x y^2 + 6 \cdot x^2 y + 3\cdot x^3$$
$$= 3x( y^2 + 2 \cdot x y + \cdot x^2)$$
$$= 3x(x+y)^2 = 3x$$

Now try it with n=4.

[edit] Another helpful trick when dealing with the general n sum. Where you have n-r you can rewrite that as n-r + 1 - 1 and regroup the 1's...

For n=4 :-

$$\sum_{r=0}^4 r\cdot C_r^4 \cdot x^r y^{4-r}$$
$$= 0 + 4 \cdot x y^3 + 12 \cdot x^2 y^2 + 12\cdot x^3 y + 4 \cdot x^4$$
$$= 4x( y^3 + 3 \cdot x y^2 + 3 \cdot x^2 y + x^3)$$
$$= 4x(x+y)^3 = 4x$$ <--------(substituting the value x+y=1)

$$= nx$$ <--------(when r=1 to n) {Is this step correct?}

Thanks for your help.

ok so far … but don't go beserk …

(and forget the ∑ … I didn't ask about the ∑)

now what else can you cancel?

and then what is the result equal to? ​

I think jambaugh's hint solves my question, so is the following required :

$$=\sum_{r=0}^{n} \frac{r}{nx} \frac{n!}{r(r-1)!(n-r)!} x^r y^{n-r}$$

I mean will this yield the same result and do we need to solve just the general term or substitute values of n like jambaugh has done.

 

[tiny-tim]

… I think jambaugh's hint solves my question, so is the following required :

$$=\sum_{r=0}^{n} \frac{r}{nx} \frac{n!}{r(r-1)!(n-r)!} x^r y^{n-r}$$

I mean will this yield the same result and do we need to solve just the general term or substitute values of n like jambaugh has done.

You're still missing the point that jambaugh and I are trying to lead you towards …

Hint: what is n!/n ?

 

[gaganspidey]

You're still missing the point that jambaugh and I are trying to lead you towards …

Hint: what is n!/n ?

This is what I've been able to work out :

$$\sum_{r=0}^{n} r C^n_r x^r y^{n-r}$$ <--------(given)

$$= \frac{n!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r}$$ <---------(cancelling the r's & adding & subtracting 1 & regrouping them)

$$= \frac{(nx)}{x} \frac{(n-1)!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r}$$ <-------(separating n & multiplying & dividing by x)

$$= (nx) C^{n-1}_{r-1} x^{r-1}y^{n-r}$$ <-------( The R.H.S)

$$= nx(y+x)^{n-1}$$ <--------(The L.H.S)

$$= nx$$ <--------(Substuting the value x+y=1)I think something's not right above, but what is it ?

 

[tiny-tim]

No, that proof is exactly right!

 

[gaganspidey]

That means I am through with this problem though I need to increase my speed a bit else...

 

[iSplicer]

Thanks, this ended up helping a lot =)

and sorry for digging up an old thread!

 

[iSplicer]

This is what I've been able to work out :

$$\sum_{r=0}^{n} r C^n_r x^r y^{n-r}$$ <--------(given)

$$= \frac{n!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r}$$ <---------(cancelling the r's & adding & subtracting 1 & regrouping them)

$$= \frac{(nx)}{x} \frac{(n-1)!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r}$$ <-------(separating n & multiplying & dividing by x)

$$= (nx) C^{n-1}_{r-1} x^{r-1}y^{n-r}$$ <-------( The R.H.S)

$$= nx(y+x)^{n-1}$$ <--------(The L.H.S)

$$= nx$$ <--------(Substuting the value x+y=1)


I think something's not right above, but what is it ?

Hmm, just a question: The solution seems impeccable, but what happened to the sigma notation after the first line?