1205 Binomial Expansion Theorem -- Sage

for n in range(5): expand((x+1)^n)

 $\newcommand{\Bold}[1]{\mathbf{#1}}1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}x + 1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{2} + 2 \, x + 1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{3} + 3 \, x^{2} + 3 \, x + 1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1$

for n in range(10): factorial(n)

 $\newcommand{\Bold}[1]{\mathbf{#1}}1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}2$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}6$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}24$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}120$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}720$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}5040$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}40320$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}362880$

#p851(45) (1/4+3/4)^5

 $\newcommand{\Bold}[1]{\mathbf{#1}}1$

def combo(n,r): return factorial(n)/factorial(r)/factorial(n-r) combo(5,3)

 $\newcommand{\Bold}[1]{\mathbf{#1}}10$

(100*combo(5,0)*(1/4)^5*(3/4)^0).n()

 $\newcommand{\Bold}[1]{\mathbf{#1}}0.0976562500000000$

for r in range(5): show((100*combo(5,r)*(1/4)^(5-r)*(3/4)^r).n())

 $\newcommand{\Bold}[1]{\mathbf{#1}}0.0976562500000000$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}1.46484375000000$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}8.78906250000000$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}26.3671875000000$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}39.5507812500000$

#10) 60 C 20 combo(60,20)

 $\newcommand{\Bold}[1]{\mathbf{#1}}4191844505805495$

#20) (x^2-y^2)^6 var('y') expand((x^2-y^2)^6)

 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{12} - 6 \, x^{10} y^{2} + 15 \, x^{8} y^{4} - 20 \, x^{6} y^{6} + 15 \, x^{4} y^{8} - 6 \, x^{2} y^{10} + y^{12}$

combo(6,0)*(x^2)^6*(-y^2)^0

 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{12}$

combo(6,1)*(x^2)^5*(-y^2)^1

 $\newcommand{\Bold}[1]{\mathbf{#1}}-6 \, x^{10} y^{2}$

combo(6,2)*(x^2)^4*(-y^2)^2

 $\newcommand{\Bold}[1]{\mathbf{#1}}15 \, x^{8} y^{4}$

combo(6,3)*(x^2)^3*(-y^2)^3

 $\newcommand{\Bold}[1]{\mathbf{#1}}-20 \, x^{6} y^{6}$

combo(6,4)*(x^2)^2*(-y^2)^4

 $\newcommand{\Bold}[1]{\mathbf{#1}}15 \, x^{4} y^{8}$

combo(6,5)*(x^2)^1*(-y^2)^5

 $\newcommand{\Bold}[1]{\mathbf{#1}}-6 \, x^{2} y^{10}$

combo(6,6)*(x^2)^0*(-y^2)^6

 $\newcommand{\Bold}[1]{\mathbf{#1}}y^{12}$

for r in range(7): combo(6,r)

 $\newcommand{\Bold}[1]{\mathbf{#1}}1$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}6$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}15$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}20$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}15$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}6$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}1$

#30) expand((2*x-3)^9)

 $\newcommand{\Bold}[1]{\mathbf{#1}}512 \, x^{9} - 6912 \, x^{8} + 41472 \, x^{7} - 145152 \, x^{6} + 326592 \, x^{5} - 489888 \, x^{4} + 489888 \, x^{3} - 314928 \, x^{2} + 118098 \, x - 19683$

combo(9,7)*(2*x)^2*(-3)^7

 $\newcommand{\Bold}[1]{\mathbf{#1}}-314928 \, x^{2}$

combo(9,5)*(2*x)^4*(-3)^5

 $\newcommand{\Bold}[1]{\mathbf{#1}}-489888 \, x^{4}$

1205 Binomial Expansion Theorem

3183 days ago by MATH4R2012