Question 1008641
<pre>
For k = 1, a<sub>1</sub> =   6 =   1+5 = 4^0+5 
For k = 2, a<sub>2</sub> =   9 =   4+5 = 4^1+5
For k = 3, a<sub>3</sub> =  21 =  16+5 = 4^2+5
For k = 4, a<sub>4</sub> =  69 =  64+5 = 4^3+5
For k = 5, a<sub>5</sub> = 261 = 256+5 = 4^4+5
...
For k = n-1, a<sub>n</sub> = 261 = 4^(n-1)+5

So we add the sequence {{{4^0+4^1+4^2+4^3+4^3+""*""*""*""+4^n}}}

with the geometric series formula to n terms:

{{{S[n]}}}{{{""=""}}}{{{(a[1](r^n-1))/(r-1)}}} with a<sub>1</sub> = 1
and r = 4.

{{{S[n]}}}{{{""=""}}}{{{(1(4^n-1))/(4-1)}}}

{{{S[n]}}}{{{""=""}}}{{{(1(4^n-1))/3}}}

{{{S[n]}}}{{{""=""}}}{{{(4^n-1)/3}}}
 
Then we must add 5 to that n times which is the same as adding 5n

Answer:  {{{SUM[n]}}}{{{""=""}}}{{{(4^n-1)/3^""+5n}}}

Edwin</pre>