(x+1)^{0} =

*1*

(x+1)^{1} =

1x + 1 =

*1 1*

(x+1)^{2} =

(x+1)(x+1) =

x*x + x*1 + 1*x + 1*1 =

1x^{2} + 2x + 1 =

*1 2 1*

(x+1)^{3} =

(x+1)(x+1)(x+1) =

((x+1)(x+1))(x+1) =

(1x^{2} + 2x + 1)(x+1) =

(1x^{2}*x + 1x^{2})+(2x*x + 2x*1)+(1*x + 1*1)=

1x^{3} + 1x^{2} + 2x^{2} + 2x + x + 1 =

1x^{3} + 3x^{2} + 3x + 1 =

*1 3 3 1*

(x+1)^{4} =

(x+1)(x+1)(x+1)(x+1) =

((x+1)(x+1)(x+1))(x+1) =

(1x^{3} + 3x^{2} + 3x + 1)(x+1) =

(1x^{3}*x + 1x^{3}*1)+(3x^{2}*x + 3x^{2}*1)+(3x*x + 3x*1)+(1*x + 1*1) =

1x^{4} + 1x^{3} + 3x^{3} + 3x^{2} + 3x^{2} + 3x + 1x + 1 =

1x^{4} + 4x^{3} + 6x^{2} + 4x + 1 =

*1 4 6 4 1*

## 12 February 2007

### Pascal and Binomial Coefficients

I realized that I never really showed how Pascal's Triangle is related to the Binomial Coefficients... So here we go:

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