How do you find the third term of #(x+3)^12#?
2 Answers
Explanation:
Any binomial expansion of the type
Hence
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Hence
Explanation:
#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#
where
In our example
The first, second and third terms of
#((12),(0))x^12 = x^12#
#((12),(1))3x^11 = 12/1*3x^11 = 36x^11#
#((12),(2))3^2x^10 = (12*11)/(2*1)*9x^10 = 66*9x^10 = 594x^10#
If you were looking for more or later terms in the expansion, then it might be easier to pick them out from the appropriate row of Pascal's triangle. Some people call the first row of Pascal's triangle the "
The terms on this row of Pascal's triangle are:
#((12),(0))# ,#((12),(1))# ,#((12),(2))# , ...#((12),(12))# .
So for example, the middle term of