Mathematics
OpenStudy (anonymous):

how do I expand and simplify the first 3 terms of: (3y + 5)^9 and (3b^2 - 2/b)^14?

OpenStudy (king):

for 3y+5)^9 use pascals triangle

OpenStudy (anonymous):

it's still the same thing? But I'm going to be using pascal's triangle for both of these binomial powers..

OpenStudy (king):

ya fr both...

OpenStudy (anonymous):

but I kind of don't get (3b^2 - 2/b)^14.. o.o

OpenStudy (king):

take 1 as x and the other as y and substitute

OpenStudy (anonymous):

huh?

OpenStudy (anonymous):

Instead of using Pascal's triangle, you can use your scientific calculator. Look for the function that has "nCr" or "xCy" on it, or anything of the like that has the "C" in the middle. And then, with that, you can use the binomial expansion formula. Let's take this as an example: $(a+b)^5 = a^5 + 5C1(a^4)(b^1) + 5C2(a^3)(b^2) + 5C3(a^2)(b^3)+5C4(a^1)(b^4)+b^5$ Wherein 5C1 is what you press on your calculator. Basically, the "a's" power decreases, and the b's powers increases. That's the easy way to put it.

OpenStudy (anonymous):

where do I find that button? o.O

OpenStudy (anonymous):

$(3y + 5)^9 = (3y)^9 + 9(3y)^8(5)^1+36(3y)^7(5)^2+84(3y)^6(5)^3...$ Do you have a scientific calculator?

OpenStudy (anonymous):

yes, I do.

OpenStudy (anonymous):

isn't y supposed increasing not decreasing?

OpenStudy (anonymous):

Hold up, let me get my textbook. I just learned this stuff today, too.

OpenStudy (anonymous):

oh xD

OpenStudy (anonymous):

Number on the left's power decreases, number on right's power increases.

OpenStudy (anonymous):

you don't need a calculator for first 3 terms $\dbinom{9}{0}=1$ $\dbinom{9}{1}=9$ $\dbinom{9}{2}=\frac{9\times 8}{2}=9\times 4=36$ so you know the first three coefficients will be 1,9,36 then $1\times (3y)^9+9\times (3y)^8\times 5^1+36\times (3y)^7\times 5^2$ are the first three terms, clean up with some arithmetic

OpenStudy (anonymous):

Oh. OH. I misread the question. The first three terms... right.

OpenStudy (anonymous):

for the exponent of 14 it is $\dbinom{14}{0}=1,\dbinom{14}{1}=14, \dbinom{14}{2}=\frac{14\times 13}{2}=7\times 13=91$ only the last one requires any real computation