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

• RChristenk
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.

RChristenk said:
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

FactChecker
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.

topsquark and WWGD
Mark44 said:
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.

topsquark

## 1. What is the binomial expansion?

The binomial expansion is a way of expanding expressions that are raised to a power, specifically expressions of the form $$(a + b)^n$$. It is expressed as a sum involving terms of the form $$\binom{n}{k} a^{n-k} b^k$$, where $$\binom{n}{k}$$ is a binomial coefficient.

## 2. What does $$(n-r+1)^{th}$$ term represent in binomial expansion?

In the context of binomial expansion, $$(n-r+1)^{th}$$ term refers to the specific term in the expanded series. If the expansion is written as a series of terms from $$T_0$$ to $$T_n$$, the $$(n-r+1)^{th}$$ term is the term that appears in the position $$n-r+1$$ in this sequence.

## 3. How do you derive the $$(n-r+1)^{th}$$ term in binomial expansion?

The $$(n-r+1)^{th}$$ term in the binomial expansion of $$(a+b)^n$$ can be derived using the general term formula $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$. For the $$(n-r+1)^{th}$$ term, set $$k = n-r+1-1 = n-r$$. Therefore, the term is $$\binom{n}{n-r} a^r b^{n-r}$$.

## 4. Why is the $$(n-r+1)^{th}$$ term important in binomial expansion?

The $$(n-r+1)^{th}$$ term is important because it allows us to identify and compute any specific term in the binomial expansion without having to expand the entire expression. This can be particularly useful for large values of $$n$$ or when only a particular term is of interest.

## 5. Can you provide an example of finding the $$(n-r+1)^{th}$$ term in a binomial expansion?

Sure! Consider the binomial expansion of $$(1+x)^5$$. To find the $$(5-2+1)^{th}$$ term, simplify to find the $$4^{th}$$ term. Using the general term formula $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$, set $$k = 4-1 = 3$$. Thus, the term is \(\binom{5}{3} 1^{5-3

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