Answer:
32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5
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Explanation:
Let's use Pascal's Triangle
In the row that starts with 1,5,... we have the values 1,5,10,10,5,1
These will be the coefficients of the terms.
Let x = 2a and y = 2b
We want to expand out (x+y)^5
Using pascals triangle, we get the following expansion
(x+y)^5 = 1x^5y^0+5x^4y^1+10x^3y^2+10x^2y^3+5x^1y^4+1x^0y^5
The numbers in bold are the coefficients 1,5,10,10,5,1 found earlier.
Note how the exponents for x start at 5 and count down to 0; while the y exponents start at 0 and count up to 5. For any term, the x and y exponents always add to 5 for the expansion of (x+y)^5. In general, the exponents of any term will add to n for (x+y)^n.
At this point, we plug in x = 2a and y = 2b
Since this will clutter things a bit, I'll do it term by term
- 1x^5y^0 = 1(2a)^5(2b)^0 = 1(32a^5)(1) = 32a^5
- 5x^4y^1 = 5(2a)^4(2b)^1 = 5(16a^4)(2b) = 160a^4b
- 10x^3y^2 = 10(2a)^3(2b)^2 = 10(8a^3)(4b^2) = 320a^3b^2
- 10x^2y^3 = 10(2a)^2(2b)^3 = 10(4a^2)(8b^3) = 320a^2b^3
- 5x^1y^4 = 5(2a)^1(2b)^4 = 5(2a)(16b^4) = 160ab^4
- 1x^0y^5 = 1(2a)^0(2b)^5 = 1(1)(32b^5) = 32b^5
So in the end, the expression (2a+2b)^5 expands out to
32a^5+160a^4b+320a^3b^2+320a^2b^3+160ab^4+32b^5
The binomial theorem uses the same basic idea, but instead of using Pascal's Triangle to get the coefficients, you'll use the nCr combination formula.
