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**