anonymous
  • anonymous
how do I expand and simplify the first 3 terms of: (3y + 5)^9 and (3b^2 - 2/b)^14?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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King
  • King
for 3y+5)^9 use pascals triangle
anonymous
  • anonymous
it's still the same thing? But I'm going to be using pascal's triangle for both of these binomial powers..
King
  • King
ya fr both...

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anonymous
  • anonymous
but I kind of don't get (3b^2 - 2/b)^14.. o.o
King
  • King
take 1 as x and the other as y and substitute
anonymous
  • anonymous
huh?
anonymous
  • 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.
anonymous
  • anonymous
where do I find that button? o.O
anonymous
  • 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?
anonymous
  • anonymous
yes, I do.
anonymous
  • anonymous
isn't y supposed increasing not decreasing?
anonymous
  • anonymous
Hold up, let me get my textbook. I just learned this stuff today, too.
anonymous
  • anonymous
oh xD
anonymous
  • anonymous
Number on the left's power decreases, number on right's power increases.
anonymous
  • 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
anonymous
  • anonymous
Oh. OH. I misread the question. The first three terms... right.
anonymous
  • 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

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