In algebra binomial expansion is the expansion of powers of a binomial. A binomial expansion is an expression in which it contains two terms eg, (a+b). This expression could also have a power on the outside of the brackets.
To generate a formula for finding the general expanded form of binomial expressions of the form (a+b)n.
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Basic Binomial Expansions
(a+b)1 = a+b
(a+b)2 = a2+2ab+ b2
(a+b)3 = a3+ 3a2b + 3ab2 + b3
(a+b)4 = a4+ 4a3b + 6a2b2+ 4ab3+ b4
The power (n) and the number of terms in each expansion is equal to the amount of terms in each expansion plus one. The coefficients in each binomial expansion is a line on Pascal’s triangle. The nth power is the number line the coefficients are on.
(a+b)1 = (1)a+b
(a+b)2 = (1)a2 + (2)ab + (1)b2
(a+b)3 = (1)a3+ (3)a2b + (3)ab2 + (1)b3
(a+b)4 = (1)a4+ (4)a3b + (6)a2b2+ (4)ab3+ (1)b4
If one was to predict the coefficients in (a+b)7 it would be the seventh line on Pascal’s triangle. The coefficients are 1, 7, 21, 35, 35, 21, 7, 1. The indices on a and b both have their pattern. Notice how the indices for a on (a+b)4 go 4,3,2,1,0 and the indices on b go 0,1,2,3,4. This pattern can be seen in any (a+b)n form. The n in the expression represents what the power for a and b would start and go down to or vice versa.
If one was to fully expand (a+b)15 and to find all coefficients, nCr needs to be used (where n is the indice on the equation, and r is a number between 0 and n). If one was to do this manually, the line the equation lies on is the nth term (in this case it lies on the 15th line) The first expression using nCr would be 15C0 which equals 1. Further expansion using nCr is shown below.
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...iven equation. A pattern in the indices was also found, this added to the efficiency of the expansion. If all these factors are brought together, a formula has been discovered that finds the general expanded form of .
To further this investigation, one could look into a formula that works for negative indices and compare this to positive integer formula. A pattern could be found from the two results and a formula could be created that works and generates answers for both positive and negative indices.
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