# help with maths a level question - binomial expansion

can anyone help with this question? :
it is given that the binomial expansion of (2+x)^5 is
32+80x+80x^2+40x^3...
hence find the coefficient of y^3 in the expansion of (2+3y+y^2)
assuming you meant to expansion of (2+3y+y^2)^5

in which case let x = 3y + y^2

only the x^2 and x^3 terms will contribute to the y^3 coefficient, specifically

x^2 = ... + 6y^3 + ...
x^3 = ... + 27y^3 + ...

so 80*6 + 40*27 = 1560
Original post by user657
can anyone help with this question? :
it is given that the binomial expansion of (2+x)^5 is
32+80x+80x^2+40x^3...
hence find the coefficient of y^3 in the expansion of (2+3y+y^2)

To find the coefficient of y^3 in the expansion of (2+3y+y^2)^5, we can use the binomial theorem:

(2+3y+y^2)^5 = (2+y)^5 + 5(2+y)^4(3y+y^2) + 10(2+y)^3(3y+y^2)^2 + 10(2+y)^2(3y+y^2)^3 + 5(2+y)(3y+y^2)^4 + (3y+y^2)^5

Since the coefficient of y^3 in the expansion of (2+x)^5 is 40, we can substitute x with 3y+y^2 to get:

(2+3y+y^2)^5 = (2+y)^5 + 5(2+y)^4(3y+y^2) + 10(2+y)^3(3y+y^2)^2 + 40(2+y)^2(3y+y^2)^3 + 5(2+y)(3y+y^2)^4 + (3y+y^2)^5

The coefficient of y^3 in this expansion is 40, so the answer is 40.

I hope this helps! Let me know if you have any other questions, and I'll gladly help
Original post by Curious_Bilawi
To find the coefficient of y^3 in the expansion of (2+3y+y^2)^5, we can use the binomial theorem:

(2+3y+y^2)^5 = (2+y)^5 + 5(2+y)^4(3y+y^2) + 10(2+y)^3(3y+y^2)^2 + 10(2+y)^2(3y+y^2)^3 + 5(2+y)(3y+y^2)^4 + (3y+y^2)^5

Since the coefficient of y^3 in the expansion of (2+x)^5 is 40, we can substitute x with 3y+y^2 to get:

(2+3y+y^2)^5 = (2+y)^5 + 5(2+y)^4(3y+y^2) + 10(2+y)^3(3y+y^2)^2 + 40(2+y)^2(3y+y^2)^3 + 5(2+y)(3y+y^2)^4 + (3y+y^2)^5

The coefficient of y^3 in this expansion is 40, so the answer is 40.

I hope this helps! Let me know if you have any other questions, and I'll gladly help

Thanks for trying to help the OP. However, I'm afraid you've over-complicated things and got the wrong answer. You are meant to use the expansion of (2+x)^5 which is given in the question.

Check Wolfram alpha. You'll find the full expansion of (2+3y+y^2)^5 is:

y^10 + 15 y^9 + 100 y^8 + 390 y^7 + 985 y^6 + 1683 y^5 + 1970 y^4 + 1560 y^3 + 800 y^2 + 240 y + 32

As you can see, the coefficient of y^3 is 1560. Luckily, calculating the full expansion is unnecessary as explained in my answer.

Hopefully this helps you with your maths too!
(2 + 3y + y^2) =
y^10 + 15 y^9 + 100 y^8 + 390 y^7 + 985 y^6 + 1683 y^5 + 1970 y^4 + 1560 y^3 + 800 y^2 + 240 y + 32

So the coefficient of y^3 is 1560, although calculating the full expansion is unnecessary.
Original post by Curious_Bilawi
To find the coefficient of y^3 in the expansion of (2+3y+y^2)^5, we can use the binomial theorem:

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