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

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The discussion focuses on finding the r-th term from both the beginning and the end of the binomial expansion of (a + 2x)^n. The coefficient for the r-th term from the beginning is expressed as nCr-1 * a^(n-r+1) * (2x)^(r-1). For the r-th term from the end, it is noted that there are n + 1 terms, and the coefficient is given as nCr-1 * a^(r-1) * (2x)^(n-r+1). Participants also clarify the correct notation for binomial coefficients, emphasizing the use of LaTeX formatting. The conversation highlights the importance of accurate algebraic representation in binomial expansions.
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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)!}##
 
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Yes, ##^nC_{n-r+1}=\dfrac{n(n-1)...(n-r+2)}{(r-1)!}##.

##n!=n(n-1)...(n-r+2)(n-r+1)!##
 
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.
 
Check your algebra in this line (one before the last in the OP): $$\dfrac{n!}{(n-r+1)!(n-(n-r+1))!}=\dfrac{n!}{(n-r+1)!(n+r-1)!}$$
 
WWGD said:
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}##
 
Mark44 said:
##\binom{n}{i}##
##\binom{n}{i}##
Where's the C?
 
PeroK said:
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.
 
RChristenk said:
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.
 
Mark44 said:
##\binom{n}{i}##
##\binom{n}{i}##
Thanks, how do you disable the editing so I can see the original Latex code?
 
  • #10
PeroK said:
Where's the C?
I think there isn't any, you just have a ##\binom {n}{i}## symbol.
 
  • #11
WWGD said:
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.
 
  • #12
PeroK said:
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.
 
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  • #13
WWGD said:
Mark44 said:
##\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.
 
  • #14
SammyS said:
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.
 
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