Question 1013237
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In the expansion of (2+x)^14 multiplied by (1+2/x)^14, find the coefficient of x^12.
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Apply the binomial formula to expand {{{(2+x)^14}}} and {{{(1+2/x)^14}}}. You will have 

{{{(2+x)^14 *(1+2/x)^14}}} = {{{(sum( C[14]^i*x^i*2^(14-i), i=0,14))}}} * {{{(sum( C[14]^j*(2/x)^j*1^(14-j), j=0,14))}}}.

where {{{C[n]^m}}} are the binomial coefficients. (See the lessons 
  <A HREF=http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Combinations-.lesson>Introduction to Combinations</A>  and 
  <A HREF=http://www.algebra.com/algebra/homework/Permutations/Binomial-Theorem.lesson>Binomial Theorem, Binomial Formula, Binomial Coefficients and Binomial Expansion</A> 

in this site).

Now, when you open parentheses, cross-multiply the terms as it is required and collect the common terms, 
you will see that {{{x^12}}} comes from these and only from these terms:

{{{C[14]^14*x^14*2^0}}} * {{{C[14]^2*(2/x)^2}}} + 

+ {{{C[14]^13*x^13*2^1}}} * {{{C[14]^1*(2/x)^1}}} +

+ {{{C[14]^12*x^12*2^2}}} * {{{C[14]^0*(2/x)^0}}}.


Thus the coefficient at {{{x^12}}} will be


{{{C[14]^14*2^0}}} * {{{C[14]^2*2^2}}} + {{{C[14]^13*2^1}}} * {{{C[14]^1*2^1}}} + {{{C[14]^12*2^2}}} * {{{C[14]^0*2^0}}}.

Next substitute here  {{{2^0}}} = 1,  {{{C[14]^0}}} = 1,  {{{C[14]^14}}} = 1, calculate the rest of binomial coefficients and get the answer. 

My part is to give you the general idea and the pivotal direction.
The rest is on you.
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