find the seventh term of the binomial expansion (x^5+y^3)^9

Answer 1

One way is use Pascal's Triangle

1

1,1

1,2,1 for ( )^2 each entry is the sum of 2 numbers just above it & to the left

1,3,3,1 for ( )^3

1,4,6,4,1

1,5,10,10,5,1

1,6,15,20,15,6,1 ( )^6

17,21,35,35,21,7,1 ( )^7

1,8,28,56,70,56,28,8,1 ( )^8

1,9,36,84,126,126,84,36,9,1 for ( )^9

7th term has a coefficient of 84

so does the 4th term

(x^5+y^3)^9 expanded, has 10 terms, 7th has a coefficient of 84

10th term is (y^3)^9

9th term is 9(x^5)(y^3)^8

8th term is 36(x^5)^2(y^3)^7

7th term is 84(x^5)^3(y^3)^6 = 84(x^15)(x^18)

each term is in descending order of (x^5) powers, and ascending order of (y^3) powers

starting with (x^5)^9(y^3)^0, then 9(x^5)^8(y^3)^1....for the 1st two terms

each term with a coefficient from Pascal's triangle