Could someone give me a hand at this one?

What is the coefficient of x[5th power] in the expansion of (x+3)[8th power]?

Possible solutions:

56

336

504

1512

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- February 10th 2007, 02:26 AMgretchencoefficient help
Could someone give me a hand at this one?

What is the coefficient of x[5th power] in the expansion of (x+3)[8th power]?

Possible solutions:

56

336

504

1512 - February 10th 2007, 03:05 AMearboth
- February 10th 2007, 03:12 AMticbol
For less computation re the binomial coefficients, use the Pascal's Triangle.

(x+3)^8

The x^5 term is in the 4th term of the expansion.

The 4th term, according to the Pascal Triangle, has 56 as the binomial coefficient.

So the 4th term is 56(x^5)(3^3) = 1512x^5

Therefore, the coefficient of the x^5 is 1512. -----------answer.

--------------------

By the nCr method,

The binomial coefficient of the 4th term is

8C3 = (8!) / (3!)(8-3)!

= (8!) / (3!)(5)!

= (8*7*6*) / (3*2*1)

= 56

Then, continue as above. - February 10th 2007, 05:12 AMCaptainBlack
- February 10th 2007, 05:34 AMSoroban
Hello, Gretchen!

Don't you know the Binomial Expansion?

If you don't, you shouldn't be trying this problem . . .

Quote:

What is the coefficient of x^5 in the expansion of (x + 3)^8 ?

Possible solutions: . 56 . . 336 . . 504 . . 1512

You should be familiar with the "combination" numbers of the expanation.

C(8,8)x^8 + C(8,7)(x^7)(3) + C(8,6)(x^6)(3^2) + C(8,5)(x^5)(3^3) + C(8,4)(x^4)(3^4) + . . .

. . . . . . . . . . . . . . . . . . . . . . 8!

So we have: . C(8,3)(3^3) .= .----- (27) .= .(56)(27) .= .1512

. . . . . . . . . . . . . . . . . . . . . .5!3!