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avatar+846 

Expand (1 + 𝑥 5 )^3 in ascending powers of 𝑥, up to and including the term in 3 x .

i know how to expand the equation but im not sure what the question is asking me to do 

 Nov 10, 2018
 #1
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+1

What is the term in brackets? Is it (1 + x^5)^3, or (1 + 5x)^3 ???

 Nov 10, 2018
 #2
avatar+846 
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(1+x^5)^3

YEEEEEET  Nov 10, 2018
 #3
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+1

Expand the following:
(x^5 + 1)^3

Expand (x^5 + 1)^3 using the binomial expansion theorem.
(x^5 + 1)^3 | = | sum_(k=0)^3 binomial(3, k) (x^5)^(3 - k) 1^k | 
 | = | binomial(3, 0) (x^5)^3 1^0 + binomial(3, 1) (x^5)^2 1^1 + binomial(3, 2) (x^5)^1 1^2 + binomial(3, 3) (x^5)^0 1^3 | : invisible comma 
binomial(3, 0) x^15 + binomial(3, 1) x^10 + binomial(3, 2) x^5 + binomial(3, 3)

Evaluate the binomial coefficients by looking at Pascal's triangle.
The binomial coeffients comprise the 4^th row of Pascal's triangle:

1 | | 3 | | 3 | | 1
(x^5)^3 + 3 (x^5)^2 + 3 x^5 + 1

For all positive integer exponents (a^n)^m = a^(m n). Apply this to (x^5)^2.
Multiply exponents. (x^5)^2 = x^(5×2):
(x^5)^3 + 3 x^(5×2) + 3 x^5 + 1

Multiply 5 and 2 together.
5×2 = 10:
(x^5)^3 + 3 x^10 + 3 x^5 + 1

For all positive integer exponents (a^n)^m = a^(m n). Apply this to (x^5)^3.
Multiply exponents. (x^5)^3 = x^(5×3):
x^(5×3) + 3 x^10 + 3 x^5 + 1Multiply 5 and 3 together.
x^15 + 3x^10 + 3x^5 + 1

 Nov 10, 2018
 #5
avatar+102466 
+1

(1 + 𝑥 5 )^3

 

\((1+x^5)^3\\ =1+3(x^5)+3(x^5)^2+(x^5)^3\\ =1+3x^5+3x^{10}+x^{15}\)

 

Normally it would be presented in reverse order but you have been asked for ascending powers of x.

Oh, I also used Pascals triangle to determine the coefficients.       1,3,3,1

 Nov 10, 2018

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