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

Answer:
The answer is option B
Step-by-step explanation:
To find the coefficient of the third term in
[tex](x + 5)^{7} [/tex]
Rewrite the expansion in the form
[tex](a + x)^{n} [/tex]
where n is the index
So we have
[tex] ({5 + x})^{7} [/tex]
After that we use the formula
[tex]nCr( {a}^{n - r} ) {x}^{r} [/tex]
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
[tex]7C2[/tex]
Hope this helps you
first of all, the notation is wrong it should be [tex] {}^nC_r \text{ and more usual notation is } {n \choose k} [/tex]
second, the
[tex](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[/tex]
here [tex] a=5 \text{ and } n=7 \text{ and for } 3^{\text{rd}} \text{ term } T_3, \quad r+1=3 \implies r=2 [/tex]
so the coefficient of third term is, [tex]{7 \choose 2}={7\choose 5}[/tex]
an important property of binomial coefficient you should know:
[tex] {n \choose k}={n \choose {n-k}}[/tex]
and if you interchange [tex] x \text{ and } a[/tex]
only the "order" will get reversed. i.e. the series will start from back.
another thing, the [tex] k^{\text{th}} \text{ term from beginning, is the } (n-k+2)^{\text{th}} \text{ term from behind}[/tex]