# Coefficient using pascals triangle What is the co-efficient of X2 in the expansion (1+2x)^5 using pascals triangle.   Any help would be greatly appreciated.

Let's establish some rules here.

Use n for the power of binomial (x+y)^n

Give values to n, starting by 0.

n = 0 => (x+y)^0 = 1

n = 1=> (x+y)^1 = x + y

n = 2 => (x+y)^2 = x^2 + 2xy + y^2

Take a look at the coefficients of the variables x and y. Once you've identified them, let's make a triangle. The vertex of the triangle is 1. The sides of triangle must be formed from the digit 1.

1

1     1

1    2    1

1   3    3    1

1   4    6    4    1

............................

Maybe you have questioned yourself about the source of the numbers found on the third row, the fourth row and so on. You must keep in mind that these numbers are the coefficients of (x+y)^3, (x+y)^4...

Notice that on the third row, the coefficients are 1,3,3,1. The numbers found on the margins must be 1 and the numbers 3 and 3 are obtained if the coefficients found on the second row are added: 1+2=3 and 2+1=3.

It is easy to determine the next row, for n = 4: 1, 1+3 = 4, 3+3 = 6, 3+1 = 4, 1.

For n=5=> 1, 4+1 = 5, 4+6 = 10, 6+4 = 10, 4+1 = 5, 1

Use Newton's binomial to find out the coefficients of (1+2x)^5 = 1*1^5 + 5*1^4*(2x) + 10*1^3*(2x)^2 + 10*1^2*(2x)^3 + 5*1*(2x)^4 + 1*(2x)^5

(1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5

Notice that the coefficient of x^2 is 40.

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Pascal's triangle begins:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 5 1 etc...

Consider the following powers of a binomial:

(a+b)^0=1

(a+b)^1=1a+1b

(a+b)^2=1a^2+2ab+1b^2

(a+b)^3=1a^3+3a^2b+3ab^2+1b^3

Notice that the coefficients are the numbers from a corresponding row in Pascal's triangle.

For your problem, the expansion is:

1(1)^5+5(1)^4(2x)^1+10(1)^3(2x)^2+10(1)^2(2x)^3+
5(1)(2x)^4+1(2x)^5

The term with x^2 is 10(1)^3(2x)^2=10*1*4x^2=40x^2

So the coefficient is 40.

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