Question 994082
5+10+20+...+(5×(2^n))
this is the summation as k varies from 0 to n of (5*2^k)
we use the fact that the partial sum, summation of k from 0 to n of 2^k = 2^(n+1) - 1, then
summation as k varies from 0 to n of (5*2^k) = 5*( 2^(n+1) - 1 )