Here the nth term can be written as T(n) = a + ((2^(n-1) - 1)d) where a = 1, d = 4, n - the order of the term. finding a summation of the above term will yield the required answer. ∑ T(n) = a*n + 4*(2^(n) - 1) - 4n { ∑ a = a*n, ∑ 2^(n-1) = [(2^(n) - 1) / (2 - 1)], ∑ 4 = 4*n & d = 4 } ∑ T(n) = n*(a - 4) + 4*(2^(n) - 1) ∑ T(n) = 4*(2^(n) - 1) - 3*n
The sequence below contains terms that doesn't have any perfect square factors (except 1). 2, 3, 5, 6, 7, 10, 11, 13, 14, 15...
What is the sum of the square-roots of the terms starting from 2 to the largest two-digit term?