Question

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

Answer 1

for 3y+5)^9 use pascals triangle

Answer 2

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

Answer 3

ya fr both...

Answer 4

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

Answer 5

take 1 as x and the other as y and substitute

Answer 6

huh?

Answer 7

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.

Answer 8

where do I find that button? o.O

Answer 9

\[ (3y + 5)^9 = (3y)^9 + 9(3y)^8(5)^1+36(3y)^7(5)^2+84(3y)^6(5)^3...\]

Do you have a scientific calculator?

Answer 10

yes, I do.

Answer 11

isn't y supposed increasing not decreasing?

Answer 12

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

Answer 13

oh xD

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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