Where Did ##(n−r+1)^{th}## Come From in Binomial Expansion?

[RChristenk]

Homework Statement

In the expansion of $(1+x)^n$, the coefficients of terms equidistant from the beginning and the end are equal.

Relevant Equations

Binomial Theorem

I'm having trouble with this concept:

In the expansion of $(1+x)^n$, the coefficients of terms equidistant from the beginning and the end are equal.
The coefficient of the $(r+1)^{th}$ term from the beginning is $^nC_r$. The $(r+1)^{th}$ term from the end has $n+1−(r+1)$, or $n−r$ terms before it; therefore counting from the beginning it is the $(n−r+1)^{th}$ term, and its coefficient is $^nC_{n−r}$, which is equal to $^nC_r$.

I understand this until "therefore counting from the beginning it is the $(n−r+1)^{th}$ term". Where did $(n−r+1)^{th}$ come from?

For example $(1+x)^6= x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Let $r=2$, then the $(r+1)$, or third term, has the coefficient $^6C_2=15$. This is correct since the third term is $15x^4y^2$.

From the end, the third term has $6+1-(2+1)=4$ terms before it, which is also correct (after the third term, it is these four terms: $20x^3y^3+15x^2y^4+6xy^5+y^6$).

"therefore counting from the beginning it is the $(n−r+1)^{th}$ term". Plugging in values gives the correct answer, $6-2+1=5$, but I cannot understand what $(n-r+1)$ actually means or where it is derived from. I have no intuition about this part. Thanks for the assistance.

 

[topsquark]

Homework Statement: In the expansion of $(1+x)^n$, the coefficients of terms equidistant from the beginning and the end are equal.
Relevant Equations: Binomial Theorem

I'm having trouble with this concept:

In the expansion of $(1+x)^n$, the coefficients of terms equidistant from the beginning and the end are equal.
The coefficient of the $(r+1)^{th}$ term from the beginning is $^nC_r$. The $(r+1)^{th}$ term from the end has $n+1−(r+1)$, or $n−r$ terms before it; therefore counting from the beginning it is the $(n−r+1)^{th}$ term, and its coefficient is $^nC_{n−r}$, which is equal to $^nC_r$.

I understand this until "therefore counting from the beginning it is the $(n−r+1)^{th}$ term". Where did $(n−r+1)^{th}$ come from?

For example $(1+x)^6= x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Let $r=2$, then the $(r+1)$, or third term, has the coefficient $^6C_2=15$. This is correct since the third term is $15x^4y^2$.

From the end, the third term has $6+1-(2+1)=4$ terms before it, which is also correct (after the third term, it is these four terms: $20x^3y^3+15x^2y^4+6xy^5+y^6$).

"therefore counting from the beginning it is the $(n−r+1)^{th}$ term". Plugging in values gives the correct answer, $6-2+1=5$, but I cannot understand what $(n-r+1)$ actually means or where it is derived from. I have no intuition about this part. Thanks for the assistance.

The expansion of $(1 + x)^n$ has n + 1 terms. If you look at the rth term from the beginning, then the rth term from the other end will have the same coefficient. That would be the (n+1)-r th term.

-Dan

 

[Mark44]

One can see the symmetry of the coefficients in Pascal's Triangle. Below are the coefficients of $(1 + x)^n$ for the first few values of n.
1 (n = 0)
1 1 (n = 1)
1 2 1 (n = 2)
1 3 3 1 (n = 3)
1 4 6 4 1 (n = 4)
and so on.

 

[WWGD]

One can see the symmetry of the coefficients in Pascal's Triangle. Below are the coefficients of $(1 + x)^n$ for the first few values of n.
1 (n = 0)
1 1 (n = 1)
1 2 1 (n = 2)
1 3 3 1 (n = 3)
1 4 6 4 1 (n = 4)
and so on.

And there's a recursion rule to Pascal 's Triangle that proves the symmetry.