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Please help asap!

 May 2, 2018

Best Answer 

 #1
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Expand the following:

(q + 0.3)^5

 

(q + 0.3)^5 = sum_(k=0)^5 binomial(5, k) q^(5 - k)×0.3^k = binomial(5, 0) q^5 0.3^0 + binomial(5, 1) q^4 0.3^1 + binomial(5, 2) q^3 0.3^2 + binomial(5, 3) q^2 0.3^3 + binomial(5, 4) q^1 0.3^4 + binomial(5, 5) q^0 0.3^5:

binomial(5, 0) q^5 + 0.3 binomial(5, 1) q^4 + 0.09 binomial(5, 2) q^3 + 0.027 binomial(5, 3) q^2 + 0.0081 binomial(5, 4) q + 0.00243 binomial(5, 5)

 

The binomial coeffients comprise the 6^th row of Pascal's triangle:

q^5 + 5 *0.3 q^4 + 10* 0.3^2 q^3 + 10* 0.3^3 q^2 + 5* 0.3^4 q + 0.3^5

 

q^5 + 1.5q^4 + 0.9q^3 + 0.27q^2 + 0.0405q + 0.00243

 May 2, 2018
 #1
avatar
+1
Best Answer

Expand the following:

(q + 0.3)^5

 

(q + 0.3)^5 = sum_(k=0)^5 binomial(5, k) q^(5 - k)×0.3^k = binomial(5, 0) q^5 0.3^0 + binomial(5, 1) q^4 0.3^1 + binomial(5, 2) q^3 0.3^2 + binomial(5, 3) q^2 0.3^3 + binomial(5, 4) q^1 0.3^4 + binomial(5, 5) q^0 0.3^5:

binomial(5, 0) q^5 + 0.3 binomial(5, 1) q^4 + 0.09 binomial(5, 2) q^3 + 0.027 binomial(5, 3) q^2 + 0.0081 binomial(5, 4) q + 0.00243 binomial(5, 5)

 

The binomial coeffients comprise the 6^th row of Pascal's triangle:

q^5 + 5 *0.3 q^4 + 10* 0.3^2 q^3 + 10* 0.3^3 q^2 + 5* 0.3^4 q + 0.3^5

 

q^5 + 1.5q^4 + 0.9q^3 + 0.27q^2 + 0.0405q + 0.00243

Guest May 2, 2018

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