How do you use the binomial series to expand #(1 + x)^12#?

1 Answer
Dec 15, 2015

Answer:

#1 + 12x + 66x^2 + 220x^3 + 495x^4 + 792x^5 + 924x^6 + 792x^7 + 495x^8 + 220x^9 + 66x^10 +12x^11 + x^12#

Explanation:

The Binomial Theorem is a very useful tool that gives us a way to find the coefficients of binomials raised to some power #n#. In this case, #n=12#, and the theorem says we can find the coefficients one of three ways:

  1. Pascal's Triangle: This is the easiest way, but also only useful if you have a copy of the triangle on hand. If you do, look at row 12 (remember the top is row 0) and use the coefficients from that row.

  2. The Choose Function: This method actually comes from the theorem itself. The function basically says how many ways there are to choose #r# items from a group of #n# objects where order does not matter.
    Turns out this function can predict binomial coefficients too. You can find them using the #n"C"r# function on your calculator where #n=12# and #r# is increasing from #0 -12#.

  3. Factorials: Should your calculator not have an #n"C"r# function, you can evaluate the function directly using factorials (#!#). The factorial of a whole number is that number multiply by all other whole numbers between #1# and itself. The formula for the choose function is:
    #n"C"r = (n!)/(r!(n-r)!)# where #r# and #n# have the same meaning as before.

No matter how you get the coefficients, the answer will still be the same:

#1 + 12x + 66x^2 + 220x^3 + 495x^4 + 792x^5 + 924x^6 + 792x^7 + 495x^8 + 220x^9 + 66x^10 +12x^11 + x^12#

Notice how as the power of #1# decreases (which doesn't actually do anything because #1^n = 1#), the power of #x# increases. The power of #1# is actually the value of #r# for a given coefficient. If you used the power of #x#, then the value of #r# would be #n# minus the exponent.

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