Asked by Raj
                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.
            
            
        Answers
                    Answered by
            oobleck
            
    (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 + ...
    
= 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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