Question

Can anyone please help me answer this question?

Using Binomial theorem find the first 5terms of the expansion

(3 − 5z)^14

Answer 1

`\qquad \qquad \textit{binomial theorem expansion} \\\\ \qquad \qquad (3-5z)^(14)~\hspace{4em} \begin{array}{clcl} term&coefficient&value\\ \cline{1-3}&\\ 1&+1&(3)^(14 )(-5z)^0\\ 2&+14&(3)^(13)(-5z)^1\\ 3&+91&(3)^(12)(-5z)^2\\ 4&+364&(3)^(11)(-5z)^3\\ 5&+1001&(3)^(10)(-5z)^4 \end{array} \\\\[-0.35em] ~\dotfill\\\\ 3^(14)-14(3^(13))(5z)+91(3^(12))(5z)^2-364(3^(11))(5z)^3+1001(3^(10))(5z)^4 \\\\\\ 4782969-111602610z+1209028275z^2-8060188500z^3+13299311025z^4`

I find the key to binomial expansion is mainly to get coefficient that goes next as you see in the middle of the table above.

now, if wonder how I get it sequentially, namely as I go, for example, how did I get 91?

well, (14 * 13 )/ 2

the current coeffient times the exponent of the 1st term divided by the "exponent of the 2nd term plus 1"

how did I get 1001?

well, (364 * 11) / 4

and so on.