Find the ##r^{th}## term 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 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.

 

[pasmith]

But technically $^nC_{n-r+1} \neq \dfrac{n(n-1)(n-2)...(n-r+2)}{(r-1)!}$.

Why not?

We have <br /> {}^nC_k = \frac{n(n-1) \cdots (n - (k - 1))}{k!}. Setting k = n - r + 1 gives <br /> \begin{split}<br /> {}^nC_{n-r+1} &amp;= \frac{n(n-1) \cdots (n - (n - r + 1 - 1))}{(n - r + 1)!} \\<br /> &amp;= \frac{n(n-1) \cdots r}{(n - r + 1)!} \\<br /> &amp;= \frac{n!}{(r-1)!(n-r+1)!} \\<br /> &amp;= \frac{n(n-1) \cdots (n- r + 2)}{(r-1)!} \\<br /> &amp;= {}^nC_{r-1}.\end{split}

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

You are missing a factor of (r - 1)! from the denominator.

 

[RChristenk]

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.

 

[Mark44]

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.