Question

7th term of (c-d)^10

Answer 1

$\bf (c-d)^{10}\implies \begin{array}{llll} term&coefficient&value\\ -----&-----&-----\\ 1&+1&(c)^{10}(-d)^0\\ 2&+10&(c)^9(-d)^1\\ 3&+45&(c)^8(-d)^2\\ 4&+120&(c)^7(-d)^3\\ 5&+210&(c)^6(-d)^4\\ 6&+252&(c)^5(-d)^5\\ 7&+210&(c)^4(-d)^6 \end{array}$

as you notice, the first term from the binomial, starts off with a highest exponent of 10, and then gradually goes down by 1 on every element.

the second term of the binomial, starts off with an exponent of 0, and goes up by 1 on every element.

now, to get the coefficient
the first coefficient is 1 and the second is the same as the binomial's exponent... those two are simple.... now, to get any subsequent one, we simply "get the product of the current coefficient and the exponent of the first term, and divide it by the exponent of the second term on the next element".

for example... how do we get 10 for the second element, basically 1 * 10/1

that one is simple

now how do we get say, 45 for the 3rd element?
10 * 9 / 2

how do we get 252 for the 6th element?
210 * 6 / 5

and so on, and from there, you can expand each element.