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.