Question 389734
Note that the difference between two consecutive terms {{{a[k]}}} and {{{a[k+1]}}} is an increasing geometric sequence. In general, we have the sequence:


{{{a[1] = k}}}


{{{a[2] = k + pr^0 = a[1] + pr^0}}}


{{{a[3] = k + pr^0 + pr^1 = a[2] + pr^1}}}


{{{a[4] = k + pr^0 + pr^1 + pr^2 = a[3] + pr^2}}}, etc.


It is seen that {{{a[i] = a[i-1] + pr^(i-2)}}} for all {{{i >= 2}}}. We can substitute p = 11 and r = 2 to obtain ,


{{{a[i] = a[i-1] + 11*2^(i-2)}}} where {{{a[1] = 15}}}


Since {{{a[5] = 180}}}, {{{a[6] = 180 + 11*2^4 = 356}}}


Note that a recursive sequence is defined on previous terms. It is possible to write each expression {{{a[i]}}} in terms of i and other constants, but it wouldn't be recursive. Using the sum of a geometric sequence, we can get


{{{a[i] = 15 + 11sum(2^j, j = 0, i-2) = 15 + 11(2^(i-1) - 1)}}}


Checking, we have {{{a[1] = 15}}}, {{{a[2] = 15 + 11 = 26}}}, {{{a[3] =15 + 11(3) = 48}}},etc. which is the same sequence as the recursion.