# How do you find the third term of #(x+3)^12#?

##### 2 Answers

#### Answer:

#### Explanation:

Any binomial expansion of the type

Hence

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Hence

#### Answer:

#### 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