## anonymous 4 years ago how do I expand and simplify the first 3 terms of: (3y + 5)^9 and (3b^2 - 2/b)^14?

1. anonymous

for 3y+5)^9 use pascals triangle

2. anonymous

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

3. anonymous

ya fr both...

4. anonymous

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

5. anonymous

take 1 as x and the other as y and substitute

6. anonymous

huh?

7. 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.

8. anonymous

where do I find that button? o.O

9. 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?

10. anonymous

yes, I do.

11. anonymous

isn't y supposed increasing not decreasing?

12. anonymous

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

13. anonymous

oh xD

14. anonymous

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

15. 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

16. anonymous

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

17. 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