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

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King
 5 years ago
Best ResponseYou've already chosen the best response.0for 3y+5)^9 use pascals triangle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's still the same thing? But I'm going to be using pascal's triangle for both of these binomial powers..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but I kind of don't get (3b^2  2/b)^14.. o.o

King
 5 years ago
Best ResponseYou've already chosen the best response.0take 1 as x and the other as y and substitute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Instead 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where do I find that button? o.O

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ (3y + 5)^9 = (3y)^9 + 9(3y)^8(5)^1+36(3y)^7(5)^2+84(3y)^6(5)^3...\] Do you have a scientific calculator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't y supposed increasing not decreasing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hold up, let me get my textbook. I just learned this stuff today, too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Number on the left's power decreases, number on right's power increases.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. OH. I misread the question. The first three terms... right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for 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
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