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Binomial expansion help please

I'm stuck on this question expand (1+2x)^16 up to and including the term x^3. Deduce the coefficient of x^3 in the expansion of (1+3x)(1+2x)^16

I've done the expansion and it's 1+32x+480x^2+4480x^3 but for the 2nd part I keep getting the answer for the coefficient wrong. The book said the answer is 5920. I worked it out as 5440.
Original post by osayukiigbinoba
I'm stuck on this question expand (1+2x)^16 up to and including the term x^3. Deduce the coefficient of x^3 in the expansion of (1+3x)(1+2x)^16

I've done the expansion and it's 1+32x+480x^2+4480x^3 but for the 2nd part I keep getting the answer for the coefficient wrong. The book said the answer is 5920. I worked it out as 5440.


I think you worked out 2x times 480x^2 instead of 3x times 480x^2, check that part of your working again
(1 + 3x)(1 + 32x + 480x^2 + 4480x^3)

Expand out:
1 + 32x + 480x^2 + 4480x^3 + 3x + 96x^2 + 1440x^3

Then simplify the coefficients of x^3:
4480x^3 + 1440x^3
= 5920
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yuki11
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Original post by anonymousm3
(1 + 3x)(1 + 32x + 480x^2 + 4480x^3)

Expand out:
1 + 32x + 480x^2 + 4480x^3 + 3x + 96x^2 + 1440x^3

Then simplify the coefficients of x^3:
4480x^3 + 1440x^3
= 5920


Thank you so much! But how did you get 1440x^3? I thought 3x multiplied by 4480 is 13440x^4
Original post by osayukiigbinoba
Thank you so much! But how did you get 1440x^3? I thought 3x multiplied by 4480 is 13440x^4


It is, but you want the coefficient of x^3, so you don't need 13440x^4. I just didn't write it down because you don't need it :smile: Sorry, bad habit.
Reply 5
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yuki11
OP
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Original post by anonymousm3
It is, but you want the coefficient of x^3, so you don't need 13440x^4. I just didn't write it down because you don't need it :smile: Sorry, bad habit.


Ok I understand now, thank you for your help :biggrin:

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