Question

Find the 7th term of the expansion of (3c + 2d)9.

Answer 1

$\bf (3c+2d)^9\\\\ -------------------------------\\\\ \begin{array}{crcll} term&&value\\ -----&&-----\\ 1&&(3c)^9(2d)^0\\ 2&+9&9(3c)^8(2d)^1\\ 3&+36&(3c)^7(2d)^2\\ 4&+84&(3c)^6(2d)^3\\ 5&+126&(3c)^5(2d)^4\\ 6&+126&(3c)^4(2d)^5\\ 7&+84&(3c)^3(2d)^6\\ \end{array}$

now.. if you notice, the exponent for the 1st term, is dropping on each term subsequently, start with highest, 9 in this case, and drops drops drops, till on the last term, will be 0

the exponent for the second term, starts off at 0, and goes up up and up on each term

that part is simple... now, the coefficient for them

the first one will have a coefficient of 1, so we can take a closer look at the 2nd instead

the coefficient for the second is 1* 9/ 1
(1) the coefficient of the current term, (9) the exponent of the 1st term, and (1) the exponent of the 2nd term on the next term

for example, how did we get 84 for the 4th term?   (36 * 7) / 3 = 84

and so on for all subsequent terms