# One of the terms in the expansion of (x+3)^7 is kx^5. Determine the value of k.

beckden | High School Teacher | (Level 1) Educator

Posted on

Another way of doing this is using the binomial coefficients.

`_nC_r=(n!)/(r!(n-r)!)`

To find the 7 power, 5th power we get

`_7C_5=(7!)/(5!(7-5)!)=(7!)/(5!2!)=(7*6*5*4*3*2*1)/(5*4*3*2*1*2*1)=(7*6)/2=21`

Now we need `(3)^(7-5)=3^2=9`

So the term is `21*x^5*9=189x^5`

jimneff116 | Student, Undergraduate | eNotes Newbie

Posted on

This is a binomial expansion problem.  The triangle below follows the form (a+b)^N.  In this case, N is the power the expansion(and also the row number we will use in the triangle) , a will be x, and b will be 3.

As you can see, the triangle is created by adding columns of 1's on the side and finding the middle terms by adding the two numbers above it. The numbers in the row give the coefficents before each term beginning with x^n + x^(n-1)b + x^(n-2)b^2.....

So the seventh row will be 1, 7, 21, 35, 35, 21, 7, 1

the term with x to the 5th power will be:  21(x^5)(b^2)

Since b=3, K=21(9)=  189