A geometric series with #N# terms is of the general form :
#G_n = a+ax^1+ax^2+...+ax^n#
Let #G_n = a\times(1+x^1+x^2+...+x^n) = a.S_n# where
#S_n = 1+x^1+x^2+...+x^n#;
There are two ways of writing #S_n#,
#S_n = 1+x\times(1+x^1+x^2+...+x^{n-1})#
#\qquad=1+xS_{n-1}# ...... (Eq.1)
#S_n = (1+x^1+x^2+...+x^{n-1}) + x^n=S_{n-1}+x^n# ...... (Eq.2)
Since the LHS of these two equations are the same we can equate their RHS.
#1+xS_{n-1}=S_{n-1}+x^n \qquad \rightarrow (x-1)S_{n-1} = (x^n-1)#
#S_{n-1} = (\frac{x^n-1}{x-1})#
Therefore, #S_n = (\frac{x^{n+1}-1}{x-1})#
This Problem: #a=5; \qquad x=4; \qquad n=6#
#G_6 = a.S_6=5\times(\frac{4^7-1}{4-1})=27305#.
