Find the ##r^{th}## term from beginning and end of ##(a+2x)^n##

[RChristenk]

Homework Statement

Find the ##r^{th}## term from the beginning and the ##r^{th}## term from the end of ##(a+2x)^n##.

Relevant Equations

Binomial-Theorem

##r^{th}## term counting from the beginning:

The coefficient of the ##r^{th}## term is ##r-1##

##^nC_{r-1}a^{n-(r-1)}(2x)^{r-1} = ^nC_{r-1}a^{n-r+1}(2x)^{r-1}##

This is the correct answer.

##r^{th}## term counting from the end:

There are a total of ##n+1## terms in ##(a+2x)^n##.

##(n+1-r)+1## is the ##r^{th}## term from the end being counted from the beginning.

The coefficient of the ##n-r+2## term is ##n-r+1##

##^nC_{n-r+1}a^{n-(n-r+1)}(2x)^{n-r+1} = ^nC_{n-r+1}a^{r-1}(2x)^{n-r+1}##

The answer is given as ##\dfrac{n(n-1)...(n-r+2)}{(r-1)!}a^{r-1}(2x)^{n-r+1}##

I'm not sure if ##^nC_{n-r+1}## in fraction form is correct:

##^nC_{n-r+1}=\dfrac{n!}{(n-r+1)!(n-(n-r+1))!}=\dfrac{n!}{(n-r+1)!(n+r-1)!}##

Or

##^nC_{n-r+1}=\dfrac{n(n-1)...(n-(n-r+1)+1)}{(n-r+1)!}=\dfrac{n(n-1)...(r+1)r}{(n-r+1)!}##

 

[Hill]

Yes, ##^nC_{n-r+1}=\dfrac{n(n-1)...(n-r+2)}{(r-1)!}##.

##n!=n(n-1)...(n-r+2)(n-r+1)!##

 

[WWGD]

The Binomial Theorem describes the expansion ##(a+b)^n## as a sum of terms, from the ##0-th##, to the ##n-th##, so you can just plug in ##r-1## in the theorem.
Sorry, don't know how to do the nCi, " n choose i" notation in Latex to write the full theorem here.

 

[Hill]

Check your algebra in this line (one before the last in the OP): $$\dfrac{n!}{(n-r+1)!(n-(n-r+1))!}=\dfrac{n!}{(n-r+1)!(n+r-1)!}$$

 

[Mark44]

Sorry, don't know how to do the nCi, " n choose i" notation in Latex to write the full theorem here.

##\binom{n}{i}##
##\binom{n}{i}##

 

[PeroK]

##\binom{n}{i}##
##\binom{n}{i}##

Where's the C?

 

[Mark44]

Where's the C?

What I wrote is the usual way to write the binomial coefficient, which is what I believe WWGD was asking about.

 

[PeroK]

Homework Statement: Find the ##r^{th}## term from the beginning and the ##r^{th}## term from the end of ##(a+2x)^n##.
Relevant Equations: Binomial-Theorem

##r^{th}## term counting from the beginning:

The coefficient of the ##r^{th}## term is ##r-1##

##^nC_{r-1}a^{n-(r-1)}(2x)^{r-1} = ^nC_{r-1}a^{n-r+1}(2x)^{r-1}##

This is the correct answer.

##r^{th}## term counting from the end:

... by symmetry must be the same with ##a## and ##2x## exchanged.

 

[WWGD]

##\binom{n}{i}##
##\binom{n}{i}##

Thanks, how do you disable the editing so I can see the original Latex code?

 

[WWGD]

Where's the C?

I think there isn't any, you just have a ##\binom {n}{i}## symbol.

 

[PeroK]

I think there isn't any, you just have a ##\binom {n}{i}## symbol.

Ah, okay, so something like ##^nC_i## is not possible! Understood.

 

[WWGD]

Ah, okay, so something like ##^nC_i## is not possible! Understood.

I never said it wasn't, only that Mark's suggestion/code did not generate it.

 

[SammyS]

##\binom{n}{i}##
##\binom{n}{i}##

Thanks, how do you disable the editing so I can see the original Latex code?

It seems that you are asking @Mark44 how he manages to show Latex code without having the double # characters invoke Latex.

He uses BB code (See the Latex guide) and changes the text color to Black for one of each pair of # characters. Of course, the other characters are Black by default, so you don't see any difference.

 

[Mark44]

He uses BB code (See the Latex guide) and changes the text color to Black for one of each pair of # characters. Of course, the other characters are Black by default, so you don't see any difference.

Yes, exactly.