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(3b + 1/2)^5

 Jul 15, 2015

Best Answer 

 #3
avatar+23254 
+5

If you know combinations:

= C(5,0)(3b)5(0.5)0 + C(5,1)(3b)4(0.5)1 + C(5,2)(3b)3(0.5)2 + C(5,3)(3b)2(0.5)3 

                                                                  + C(5,4)(3b)1(0.5)4 + C(5,5)(3b)0(0.5)5 

=  1(3b)5(0.5)0 + 5(3b)4(0.5)+ 10(3b)3(0.5)2 + 10(3b)2(0.5)3 

                                                                  + 5(3b)1(0.5)4 + 1(3b)0(0.5)5 

=  1(243b5)(1) + 5(81b4)(0.5) + 10(27b3)(0.25) + 10(9b2)(0.125)

                                                                  + 5(3b)(0.0625) + 1(1)(0.03125)

=  243b3 + 202.5b4 + 67.5b3 + 11.25b2 + 0.9375b + 0.03125

If you don't know the rule of combinations, use the 5th row of Pascal's triangle.

 Jul 15, 2015
 #1
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0

here you go buddy the answer is 3b raised to power of 5   +0.03125

 Jul 15, 2015
 #2
avatar+118696 
+5

Sorry anon - That is not correct.

you need to use the binomial expansion

I am on my phone at the moment so it is all to difficult for me right now.

if no one gives a timely answer why not Google  binomial expansion and watch a you tube presentation.

it would most likely be easier to understand anyway. :)

 Jul 15, 2015
 #3
avatar+23254 
+5
Best Answer

If you know combinations:

= C(5,0)(3b)5(0.5)0 + C(5,1)(3b)4(0.5)1 + C(5,2)(3b)3(0.5)2 + C(5,3)(3b)2(0.5)3 

                                                                  + C(5,4)(3b)1(0.5)4 + C(5,5)(3b)0(0.5)5 

=  1(3b)5(0.5)0 + 5(3b)4(0.5)+ 10(3b)3(0.5)2 + 10(3b)2(0.5)3 

                                                                  + 5(3b)1(0.5)4 + 1(3b)0(0.5)5 

=  1(243b5)(1) + 5(81b4)(0.5) + 10(27b3)(0.25) + 10(9b2)(0.125)

                                                                  + 5(3b)(0.0625) + 1(1)(0.03125)

=  243b3 + 202.5b4 + 67.5b3 + 11.25b2 + 0.9375b + 0.03125

If you don't know the rule of combinations, use the 5th row of Pascal's triangle.

geno3141 Jul 15, 2015

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