Question

I need some help with binomial theorems. can anyone help solve this?

find the seventh term of (3x + 2y)^9

Answer 1

`\bf (3x+2y)^9\\\\ -----------------------------\\\\ \begin{array}{crllll} term&value\\ -----&------\\ 1&1(3x)^9(2y)^0\\\\ 2&9(3x)^8(2y)^1\\\\ 3&36(3x)^7(2y)^2\\\\ 4&84(3x)^6(2y)^3\\\\ 5&126(3x)^5(2y)^4\\\\ 6&126(3x)^4(2y)^5\\\\ 7&84(3x)^3(2y)^6 \end{array}`

now, how the heck do we get the coefficient for the term, those 84, 126, 9 and so on?

well, you grab the current coefficient, multiply it times the exponent of the first component,
whatever product you get,
divide it by the exponent of the second component in the "next" term
which will be, whatever the current is +1

for example, how the heck did we get 36 for the 3rd term?

well, the second term has 9 as coefficient, the first component on the second term has an exponent of 8
9 * 8 = 72
the second component, has an exponent of 1, we add one 1 + 1, is the next exponent of it, or the exponent on the next term, thus

(9 * 8)/2 = 36

Answer 2

Use Pascal's Triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1 (this is the row you use)

So, the seventh term of (3x + 2y)^9 will be 84xy