Binomial Theorem

[Yunie]

Write down and simplify the first four terms in the expansion of [1- (x/2)]^10.
Hence find the coefficient of x^3 in the expansion of (5+4x)(1 - x/2))^10.

My workings:

[1- (x/2)]^10 = 1 + 10C1 (-x/2) + 10C2 (-x/2)^2 + 10C3 (-x/2)^3 + ...
= 1 + 10 (-x/2) + 45 (x^2/4) + 120 (-x^3/8) + ...
= 1 - 5x + 45x^2/4 + 15x^3 + ...


(5+4x)(1-x/2)^10 = (5+4x)(1-5x+45x^2/4 + 15x^3)
= (5*15)x^3 + (4*45/x)x^3
= 75x^3 + 180x^3/4
= 75x^3 + 45x^3

coefficient of x^3 = 75+45 = 120

The answer is -30...Please help me check my workings and correct me...It would be best for you to show your workings step-by-step..Thanks.


Also, can help me do this question?:

[x-(2/x^2)]^16

Thanks.

[krtxmrtz]

The first part is almost correct: only the sign of the 4th term must be negative:

[1- (x/2)]¹⁰ = 1 - 5x + 45x²/4 - 15x³ + ...

and this is why in the second part you get 120 rather than -30, check it!

[Yunie]

Hahaha, I see. Careless! Thanks for pointing out the mistake for me krtxmrtz!

Hmm, perhaps you could give me some hints on how to do for this question>>> [x-(2/x^2)]^16 ?

Thanks a lot.

[krtxmrtz]

Hahaha, I see. Careless! Thanks for pointing out the mistake for me krtxmrtz!

Hmm, perhaps you could give me some hints on how to do for this question>>> [x-(2/x^2)]^16 ?

Thanks a lot.

Do you know & understand the binomial expansion?

[Yunie]

I know...But [x-(2/x^2)]^16 is different from [1- (x/2)]^10...

[Yunie]

And, how to find the coefficient of x in the expansion of
[x-(2/x^2)]^16?

Thanks again.

[krtxmrtz]

And, how to find the coefficient of x in the expansion of
[x-(2/x^2)]^16?

Thanks again.

The expansion of (a + b)ⁿ has n+1 terms and the j-th term is:

[n! / ((j - 1)!*(n - j + 1)!)] a^{n - j + 1}b^{j - 1}

and the sum of the powers of a and b is always n:
n - j + 1 + (j - 1) = n

In your specific case, the j-th term is:

[16! / ((j - 1)!*(17 - j)!)] x^{16 - j + 1}(-2/x²)^{j - 1} = [16! / ((j - 1)!*(17 - j)!)] x^{17 - j}(-2*x^-2)^{j - 1} = [16! / ((j - 1)!*(17 - j)!)] x^{17 - j}(-2)^{j - 1}*x^{-2(j - 1)}

so you're aiming at a j value such that the power of x is 1,

17 - j -2(j - 1) = 1 wherefore j = 6, so finally the term you want is:

[16! / ((6 - 1)!*(17 - 6)!)] x^{17 - 6}(-2)^{6 - 1}*x^{-2(6 - 1)} = [16! / (5! * 11!)]x¹¹*(-2)⁵x^-10 = -4362*32 x = -139776 x

And to be strict, the coefficient is just -139776

[Yunie]

Your workings look so complicated...*blur*

[krtxmrtz]

Your workings look so complicated...*blur*

Yes, I was trying to do it such that you didn't have to write the entire expansion up to the 6th term, but it finally turned out more complicated. You can go on with the binomial theorem and explicitely write the expansion for (a + b)ⁿ for n = 16 and then substitute the values for your particular case, i.e. a = x and b = 2 / x²

[Yunie]

Ok...I will try..Actually I used to know how to do this type of question, it just slips out of my mind when I did not touch this chapter for quite some time..It's okay, thanks for your help.