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

How would you find the coefficient of the third term in (x+5)^7?-example-1

2 Answers

2 votes

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}

User Almoraleslopez
by
7.1k points
4 votes

Answer:

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

User Andrey Marchuk
by
7.9k points

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