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

  • #1
RChristenk
49
4
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 from beginning: ##^nC_{r-1}a^{n-r+1}(2x)^{r-1}##

For the ##r^{th}## term from the end, we first know there are a total of ##n+1## terms in this binomial expansion. Subtracting the (##r^{th}## term from the end) from the total number of terms, ##n+1##, results in ##n+1-r## which represents the number of terms before it (counting from the beginning). Hence counting from the beginning, the ##r^{th}## term from the end is represented by ##n-r+2##.

Therefore ##r^{th}## term from the end: ##^nC_{n-r+1}a^{r-1}(2x)^{n-r+1}##.

The answers are given in an expanded coefficient form. For the ##r^{th}## term from the beginning: ##\dfrac{n(n-1)(n-2)...(n-r+2)}{(r-1)!}a^{n-r+1}(2x)^{r-1}##.

For the ##r^{th}## term from the end: ##\dfrac{n(n-1)(n-2)...(n-r+2)}{(r-1)!}a^{r-1}(2x)^{n-r+1}##.

At first, I didn't understand why the coefficient expansion was equivalent. Then I realized ##^nC_{r-1}=^nC_{n-r+1}##.

But technically ##^nC_{n-r+1} \neq \dfrac{n(n-1)(n-2)...(n-r+2)}{(r-1)!}##. Shouldn't it be something like ##\dfrac{n!}{(n-r+1)!}##? How would I write the full expansion of ##^nC_{n-r+1}##? Thanks.
 
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  • #2
RChristenk said:
But technically ##^nC_{n-r+1} \neq \dfrac{n(n-1)(n-2)...(n-r+2)}{(r-1)!}##.
Why not?

We have [tex]
{}^nC_k = \frac{n(n-1) \cdots (n - (k - 1))}{k!}.[/tex] Setting [itex]k = n - r + 1[/itex] gives [tex]
\begin{split}
{}^nC_{n-r+1} &= \frac{n(n-1) \cdots (n - (n - r + 1 - 1))}{(n - r + 1)!} \\
&= \frac{n(n-1) \cdots r}{(n - r + 1)!} \\
&= \frac{n!}{(r-1)!(n-r+1)!} \\
&= \frac{n(n-1) \cdots (n- r + 2)}{(r-1)!} \\
&= {}^nC_{r-1}.\end{split}[/tex]

Shouldn't it be something like ##\dfrac{n!}{(n-r+1)!}##?

You are missing a factor of [itex](r - 1)![/itex] from the denominator.
 
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  • #3
How did you go from ##\dfrac{n!}{(r-1)!(n-r+1)!}## to ##\dfrac{n(n-1)(n-2)\cdots(n-r+2)}{(r-1)!}##?

I am assuming ##n!=n(n-1)(n-2)\cdots (n-r+2)(n-r+1)(n-r)\cdots 1##. Would this be correct? Thanks.
 
  • #4
RChristenk said:
How did you go from ##\dfrac{n!}{(r-1)!(n-r+1)!}## to ##\dfrac{n(n-1)(n-2)\cdots(n-r+2)}{(r-1)!}##?

I am assuming ##n!=n(n-1)(n-2)\cdots (n-r+2)(n-r+1)(n-r)\cdots 1##. Would this be correct? Thanks.
Your assumption is correct.
 

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