- #1

RChristenk

- 63

- 9

- 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 "

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##).

"

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.