First, notice that if we add the powers for a and b the result is always equal to the original power given to the term.
For example in (a+b)^{ 2}= a ^{ 2}+2ab+b ^{2} notice that power in each term is always equal to 2. The first a^{2}, the second a ^{1} and b ^{1} and again 1+1 is equal 2. The same goes for a^{3}+3a^{2}b+3ab^{3}+b^{3} and on and on.
Go to the next coming post.
Second, it is apparent that power decreases from a, and increases in b as we go forward.
For example in (a+b)^{5}=a ^{ 5} b ^{ 0}+ a ^{ 4 }b ^{ 1} + a ^{ 3} b ^{ 2} + a ^{ 2} b ^{ 3} + a ^{1} b ^{ 4} + a ^{ 0} b ^{ 5}.
As it must become obvious from the above, a starts with power 5 and goes to power 0 and b starts with power 0 and goes to power 5. Of course in the above we are missing the coefficient for each term. Now I show you how to find them.
To make writing the coefficients clear I rewrite the above powering using only coefficients.
i.e. how many of each kind of term:
(a + b)
1 1 0+
0 1 1
-------
1 2 1 0
(a + b)^{2}
1 2 1 0+
0 1 2 1
----------
(a + b) ^{3}
1 3 3 1 0 +
0 1 3 3 1
-------------
(a + b) ^{4}
1 4 6 4 1 +
0 1 4 6 4 1
---------------------------
a + b) ^{5}
1 5 10 10 5 1
This is what is known as Pascal's Triangle. The last thing that you have to do is substitute 2x for a and y for b in the above. Good luck
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