Well if #n = 2# then there isn't a sum, the answer is just:
#10(2/3)^2 = 10(4/9) = 40/9#
But perhaps the question was meant to ask that the infinite sum be taken starting at #n=2# such that the equation is:
#sum_(n=2)^infty 10(2/3)^n#
In this case, we would compute it by first noting that any geometric series can be seen as being of the form:
#sum_(n=0)^infty ar^n#
In this case, our series has #a = 10# and #r = 2/3#.
We will also note that:
#sum_(n=0)^infty ar^n = asum_(n=0)^infty r^n #
So we can simply compute the sum of a geometric series #(2/3)^n# and then multiply that sum by #10# to arrive at our result. This makes things easier.
We also have the equation:
#sum_(n=0)^infty r^n = 1/(1-r)#
This allows us to compute the sum of the series starting from #n=0#. But we want to compute it from #n=2#. In order to do this, we will simply subtract the #n=0# and #n=1# terms from the full sum. Writing the first several terms of the sum out we can see that it looks like:
#1 + 2/3 + 4/9 + 8/27 + ...#
We can see that:
#sum_(n=2)^infty 10(2/3)^n = 10sum_(n=2)^infty (2/3)^n = 10[sum_(n=0)^infty (2/3)^n - (1 + 2/3)]#
#=10[1/(1-(2/3)) - (1 + 2/3)]#
#= 10[3 - 5/3] = 10[9/3 - 5/3] = 40/3#
