SOLUTION: I need help with this question. Which are the first four terms in the expansion (2-3y) power of 8?

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Question 715610: I need help with this question. Which are the first four terms in the expansion (2-3y) power of 8?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There is a general formula that you were probably given, but probably makes no sense to you.
If your teacher wants it, write it the way it was given to you, but I will tell you what ideas I would use to get the answer (without worrying about formulas).

Since the power is 8 , there will be 8%2B1=9 terms.
I could start with 2%5E8=256 as a first term,
and end with %28-3y%29%5E8 as a last term,
and that is probably what's expected.

All the terms will have combinations of powers of 2
and powers of %28-3y%29 with exponents adding up to 8.
The exponents of %28-3y%29 increase from term to term, starting at zero and going up to 8.
The exponents of 2 decrease from 8 to zero.
The in-between terms, between the first term and the last term,
will also have combinatorial numbers as extra factors.
(There are many ways to write the symbols for combinatorial numbers,
but I do not know what way you would use in your class,
and I can only write combinatorial numbers in the form %28matrix%282%2C1%2Cn%2Ck%29%29 on this website).
Those combinatorials are all combinations of 8
and the other number in the combination is one of the exponents.
Traditionally, it would be the exponent on %28-3y%29.
(If you used the exponent of 2 instead,
the symbol would look different,
but it would calculate as the same number).

So term number 4 would have
%28-3y%29%5E%28%284-1%29%29=%28-3y%29%5E3=-27y%5E3
with 3 as the exponent for %28-3y%29.
It would also have
2%5E%28%288-3%29%29=2%5E5=32 and
it would also have the combinatorial number
%28matrix%282%2C1%2C8%2C3%29%29=8%2A7%2A6%2F1%2F2%2F3=+56
So the fourth term is
56%2A32%2A%28-27y%5E3%29=-48384y%5E3

The third term has %28matrix%282%2C1%2C8%2C2%29%29=8%2A7%2F1%2F2=28
%28-3y%29%5E%28%283-1%29%29%29=%28-3y%29%5E2=9y%5E2 and 2%5E%28%288-2%29%29%29=2%5E6=64
So the third term is 28%2A64%2A%289y%5E2%29=16128y%5E2

The second term has %28matrix%282%2C1%2C8%2C1%29%29=8%2F1=8
%28-3y%29%5E%28%282-1%29%29%29=%28-3y%29%5E1=-3y and 2%5E%28%288-1%29%29%29=2%5E7=128
So the secondd term is 8%2A128%2A%28-3y%29=-3072y

Then the first four terms would be
256-3072y%2B16128y%5E2-48384y%5E3
If you have to "show your work", you could just write


A general formula for term number k of the expansion of %28a%2Bb%29%5En
is %28matrix%282%2C1%2Cn%2Ck%29%29%2Aa%5E%28%28n-k%2B1%29%29%2Ab%5E%28%28k-1%29%29