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Expand (2+3x)^5 in ascending power of x, up to x^3.

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merishorverde | Student, College Freshman | eNoter

Posted January 27, 2011 at 3:21 AM via web

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Expand (2+3x)^5 in ascending power of x, up to x^3.

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giorgiana1976 | College Teacher | Valedictorian

Posted January 27, 2011 at 3:27 AM (Answer #1)

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We'll apply Binomial Theorem:

(a+b)^n = C(n,0)*a^n + C(n,1)*a^(n-1)*b + ... + C(n,k)*a^(n-k)*b^k + .... + C(n,n)*b^n

We'll put a = 2, b = 3x and n = 5

(2+3x)^5 = C(5,0)*2^5 + C(5,1)*2^4*3x + C(5,2)*2^3*9x^2 + C(5,3)*2^2*27x^3 + ...

C(5,0) = 1

C(5,1) = 5

C(5,2) = 5(5-1)/2 = 10 = C(5,3)

(2+3x)^5 = 32 + 5*16*3x + 10*8*9x^2 + 10*4*27x^3 + ...

(2+3x)^5 = 32 + 240x + 720x^2 + 1080x^3 + ...

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 27, 2011 at 3:31 AM (Answer #2)

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We have to expand (2 + 3x)^5

Using the binomial theorem:

(2 + 3x)^5 = 2^5 + 5*2^4*(3x) + 10*2^3*(3x)^2 + 10*2^2*(3x)^3 + 5*2^1*(3x)^4 + (3x)^5

=> 32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5

The required result is 32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5

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