Math olympiad products question

[timetraveller123]

That n is not the same as the n in the problem.

oh you careless me but i am still getting a different result this is my working

$3^{2^{n+1}} ( 1- (\frac{2}{3})^{2^{n+1} -1 }) \\ 3^{2^{n+1}} - 3( 2^{2^{n+1} -1 })$
now what am i doing wrong

 

[haruspex]

now what am i doing wrong

You are not applying the upper bound correctly in the sum. If the sum is up to the term r^k then the numerator should have 1-r^k+1. You omitted the +1.

 

[timetraveller123]

oh wait the index starts at i=0 not 1 so it must be 2²ⁿ⁺¹ correct ?

then it is equal to
$3^{2^{n+1}} - 2^{2^{n+1}}$

 

[haruspex]

oh wait the index starts at i=0 not 1 so it must be 2²ⁿ⁺¹ correct ?

then it is equal to
$3^{2^{n+1}} - 2^{2^{n+1}}$

Yes.