Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
9. Binomial Distributions
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Exercise 3 Page 735

Recall the formula for the expansion of (a+b)^n stated by the Binomial Theorem.

20c^3d^3

Practice makes perfect

To find the 4^(th) term of the binomial expansion, we should recall the Binomial Theorem. It states that for every positive integer n, we can expand the expression (a+b)^n by using the numbers in the n^(th) row of Pascal's Triangle.

(a+b)^n = P_0 a^nb^0 + P_1 a^(n-1)b + ... + P_(n-1) ab^(n-1) + P_n a^0b^n

It can be shown that the i^\text{th} coefficient of the n^\text{th} row of the Pascal's Triangle equals _nC_i. Therefore, we can restate the binomial theorem using combinations.

(a+b)^n = _nC_0 a^nb^0 + _nC_1 a^(n-1)b + ... + _nC_(n-1) ab^(n-1) + _nC_n a^0b^n

Next, let's recall the formula to calculate the combination _nC_x. _nC_x ⇔ n!/x!(n-x)! We will use this to substitute the values of the combinations in the formula for the binomial expansion.

(a+b)^n = _nC_0 a^nb^0 + _nC_1 a^(n-1)b + ... + _nC_(n-1) ab^(n-1) + _nC_n a^0b^n
( c+ d)^6= 1 c^6 d^0+ 6 c^5 d^1+ 15 c^4 d^2+ 20 c^3 d^3+ 15 c^2 d^4+ 6 c^1 d^5+ 1 c^0 d^6
Finally, let's simplify the expression.
1c^6 d^0+6 c^5 d^1+ 15 c^4 d^2+ 20 c^3 d^3+ 15 c^2 d^4+ 6 c^1 d^5+1 c^0 d^6
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Simplify
1c^6 (1)+6 c^5 d^1+ 15 c^4 d^2+ 20 c^3 d^3+ 15 c^2 d^4+ 6 c^1 d^5+1 (1) d^6
1c^6 (1)+6 c^5 d+ 15 c^4 d^2+ 20 c^3 d^3+ 15 c^2 d^4+ 6 c d^5+1 (1) d^6
c^6+6 c^5 d+ 15 c^4 d^2+ 20 c^3 d^3+ 15 c^2 d^4+ 6 c d^5+d^6
The 4^(th) term of this expansion is 20 c^3 d^3.

Showing Our Work

Calculating _nC_x
As previously mentioned, there is a formula to calculate the value of _nC_x. _nC_x ⇔ n!/x!(n-x)! Let's calculate _6C_0 using this formula.
_6C_0
6!/0!(6-0)!
6!/0!6!
6!/0!6!
1/0!

0!=1

1/1
1
We found that _6C_0=1. By following the same procedure, the values of the other combinations can be calculated.