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For some real number a and some positive integer n, the first few terms in the expansion of (1 + ax)^n are 1 - 12x + 54x^2 + cx^3 + ... Find c.

 Jan 25, 2025
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For the second term

C(n,1) (- a)  = -12

n! / (n-1)! * a  = 12                {   n!  /(n -1)!  =  (n) }

n * a  =  12

a = 12/n    →   a^2  =  144/n^2

 

For the third term

 C (n,2) (-a)^2  = 54

n! / [ (n -2)! (2!) ] * a^2  = 54             { n! / (n -2)!  =  (n)(n -1) }

(n)(n-1) * 144/n^2 = 108

(n^2 - n) / n^2  =  108 / 144 =  3 / 4

1n^2 - n  = (3/4)n^2

(1/4)n^2 - n  =  0

n (1/4 * n - 1)  = 0

n = 0   reject

n  = 4  accept

 

Since the expansion  terms  alternate signs,  then  

 

a  =  -12/n  =  -12/4  = -3

 

The binomial is  (1 -3x)^4

 

c =  C(4, 3) * (-3)^3   =    4 *  -27  =    -108

 

cool cool cool

 Jan 25, 2025
edited by CPhill  Jan 25, 2025

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