Question 1117363
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The binomial expansion is this formula


{{{(a+b)^n}}} = {{{a^n}}} + {{{C[n]^1*a^(n-1)*b}}} + {{{C[n]^2*a^(n-2)*b^2}}} + {{{C[n]^3*a^(n-3)*b^3}}} + . . . + {{{C[n]^(n-1)*a^1*b^(n-1)}}} + {{{b^n}}}


    (see this Wikipedia article <A HREF=https://en.wikipedia.org/wiki/Binomial_theorem>https://en.wikipedia.org/wiki/Binomial_theorem</A>  or my lesson

     <A HREF=https://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A>  in this site).


For our case, the term containing {{{x^k}}} is  {{{C[19]^k*2^(19-k)*(3x)^k}}},  while the term containing {{{x^(k+1)}}} is  {{{C[19]^(k+1)*2^(19-(k+1))*(3x)^(k+1)}}},

and they want to know at which k the coefficients at  {{{x^k}}}  and  {{{x^(k+1)}}}  are the same:

    {{{C[19]^k*2^(19-k)*3^k}}} = {{{C[19]^(k+1)*2^(19-(k+1))*3^(k+1)}}}.    (1)

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


From (1), you have  

    {{{C[19]^k/C[19]^(k+1)}}} = {{{(2^19-(k+1)*3^(k+1))/(2^(19-k)*3^k)}}},   or, simplifying the right side,   {{{C[19]^k/C[19]^(k+1)}}} = {{{3/2}}}.    (2)


Next, use  {{{C[19]^k}}} = {{{19!/(k!*(19-k)!)}}},  {{{C[19]^(k+1)}}} = {{{19!/((k+1)!*(19-(k+1))!)}}}.   

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

    {{{(k+1)/(19-k)}}} = {{{3/2}}},

    2*(k+1) = 3*(19-k)  ====>  2k + 2 = 57 - 3k  ====>  2k + 3k = 57 - 2  ====>  5k = 55  ====>  k = {{{55/5}}} = 11.


<U>Answer</U>.  k = 11.
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