SOLUTION: if the coefficients of x^k and x^(k+1) in the expansion of (2+3x)^19 are equal, find k

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Question 1117363: if the coefficients of x^k and x^(k+1) in the expansion of (2+3x)^19 are equal, find k
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
The binomial expansion is this formula


%28a%2Bb%29%5En = a%5En + C%5Bn%5D%5E1%2Aa%5E%28n-1%29%2Ab + C%5Bn%5D%5E2%2Aa%5E%28n-2%29%2Ab%5E2 + C%5Bn%5D%5E3%2Aa%5E%28n-3%29%2Ab%5E3 + . . . + C%5Bn%5D%5E%28n-1%29%2Aa%5E1%2Ab%5E%28n-1%29 + b%5En


    (see this Wikipedia article https://en.wikipedia.org/wiki/Binomial_theorem  or my lesson

     Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion  in this site).


For our case, the term containing x%5Ek is  C%5B19%5D%5Ek%2A2%5E%2819-k%29%2A%283x%29%5Ek,  while the term containing x%5E%28k%2B1%29 is  C%5B19%5D%5E%28k%2B1%29%2A2%5E%2819-%28k%2B1%29%29%2A%283x%29%5E%28k%2B1%29,

and they want to know at which k the coefficients at  x%5Ek  and  x%5E%28k%2B1%29  are the same:

    C%5B19%5D%5Ek%2A2%5E%2819-k%29%2A3%5Ek = C%5B19%5D%5E%28k%2B1%29%2A2%5E%2819-%28k%2B1%29%29%2A3%5E%28k%2B1%29.    (1)

        (notice the difference between "the terms" and "the coefficients" !)


From (1), you have  

    C%5B19%5D%5Ek%2FC%5B19%5D%5E%28k%2B1%29 = %282%5E19-%28k%2B1%29%2A3%5E%28k%2B1%29%29%2F%282%5E%2819-k%29%2A3%5Ek%29,   or, simplifying the right side,   C%5B19%5D%5Ek%2FC%5B19%5D%5E%28k%2B1%29 = 3%2F2.    (2)


Next, use  C%5B19%5D%5Ek = 19%21%2F%28k%21%2A%2819-k%29%21%29,  C%5B19%5D%5E%28k%2B1%29 = 19%21%2F%28%28k%2B1%29%21%2A%2819-%28k%2B1%29%29%21%29.   

Substitute it into the left side fraction of (2),  and you will get after canceling common factors

    %28k%2B1%29%2F%2819-k%29 = 3%2F2,

    2*(k+1) = 3*(19-k)  ====>  2k + 2 = 57 - 3k  ====>  2k + 3k = 57 - 2  ====>  5k = 55  ====>  k = 55%2F5 = 11.


Answer.  k = 11.