Question

How would you find the coefficient of the third term in (x+5)^7?

Answer 1

first of all, the notation is wrong it should be
`{}^nC_r \text{ and more usual notation is } {n \choose k}`

second, the

`(r+1)^{\text{th}} \text{ term } T_(r+1) \text{ in the expansion of } (x+a)^n \text{ is } {n \choose r}x^((n-r))a^r`

here
`a=5 \text{ and } n=7 \text{ and for } 3^{\text{rd}} \text{ term } T_3, \quad r+1=3 \implies r=2`

so the coefficient of third term is,
`{7 \choose 2}={7\choose 5}`

an important property of binomial coefficient you should know:

`{n \choose k}={n \choose {n-k}}`

and if you interchange
`x \text{ and } a`

only the "order" will get reversed. i.e. the series will start from back.

another thing, the
`k^{\text{th}} \text{ term from beginning, is the } (n-k+2)^{\text{th}} \text{ term from behind}`

Answer 2

The answer is option B

Explanation:

To find the coefficient of the third term in

`(x + 5)^(7)`

Rewrite the expansion in the form

`(a + x)^(n)`

where n is the index

So we have

`({5 + x})^(7)`

After that we use the formula

`nCr( {a}^(n - r) ) {x}^(r)`

where r is the term we are looking for

For the third term we are looking for the term containing x²

that's

r + 1 = 3

r = 2

So to find the coefficient of the third term

We have

`7C2`

Hope this helps you