Question 1158664
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<pre>

In this problem, the sequence is close to be a Geometric progression, but actually, is not.


It requires an ACCURATE analysis. First time the ball falls down; then every next bouncing, it goes UP and DOWN


The total distance is the sum of these values

            Down   Up and       Up and       Up and       Up and 
                    down         down         down         down

              2     {{{2*(3/4)}}}        {{{2*(3/4)^2}}}       {{{2*(3/4)^3}}}       {{{2*(3/4)^4}}}

Coefficient   1       2            2            2           2    <<<---=== this coefficient accounts for "up and down"


So the total distance is

   2   +  2*S,   where S is the infinite geometric progression with the first term a = {{{2*(3/4)}}} = {{{6/4}}}  and the common ratio r = {{{3/4}}}.


The sum of this infinite progression is  S = {{{a/(1-r)}}} = {{{((6/4))/((1-3/4))}}} = {{{((6/4))/((1/4))}}} = 6.


Therefore, the <U>ANSWER</U>  to the problem question is 2 + 2*6 = 14 meters.
</pre>

Solved.


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So, this problem has a "trap".


Those who have experience in solving/teaching/explaining such problems, know about it.


All the others, as a rule, go directly to this trap.