Question

The first 2 values in ascending power of x in the expansion of (1+x) (2-x/4)^n are p+qx^2.Find the values of n,p,q.

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asked by Raj
today at 12:10pm
(2 - x/4)^n = 2^n - n*2^(n-1) * (x/4) + n(n-1)/2 * 2^(n-2) * (x/4)^2 - ...
= 2^n - n*2^(n-3) *x + n(n-1)*2^(n-7)*x^2 - ...
Now multiply that by 1+x and you have
2^n - n*2^(n-3) *x + n(n-1)*2^(n-7)*x^2 - ... + 2^n*x - n*2^(n-3) *x^2 + n(n-1)*2^(n-7)*x^3 - ...

= 2^n + (2^n - n*2^(n-3))*x + (n(n-1)*2^(n-7) - n*2^(n-3))*x^2 + ...
So, if you want the x term to vanish, you need
2^n - n*2^(n-3) = 0
n = 8

Now that gives you
2^8 + (8(8-1)*2^(8-7) - 8*2^(8-3))*x^2 = 256 - 144x^2 + ...

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👨‍🏫
oobleck
today at 12:42pm

Why n(n-1)/2 in third term?

Answer 1

An example

(a+b)^5
The binomial expansion gives us
C(5,0)(a^5)(b^0) + C(5,1)(a^4)(b^1) + C(5,2)(a^3)(b^2) + C(5,3)(a^2)(b^3) + C(5,4)(a^1)(b^4) + C(5,5)(a^0)(b^5)
= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5
look at the coefficients: 1 5 10 10 5 1, they are the 6th row of Pascal's triangle, I hope
that this important property has been taught to you or pointed out to you

recall how we calculate something like C(5,2)
= 5!/(2!3!) = 5*4/2
C(n,2) would be n!/(2!(n-2)!)
= n(n-1)(n-2)(n-3)...(2)(1) / ((n-2)(n-3)...(2)(1) * 2(1) )
= n(n-1)/2!

C(n,4) would be n(n-1)(n-2)(n-3)(n-4)/4!

etc