How do you find sum of geometric series 5+20+80+...+ 20480?

1 Answer
Feb 27, 2016

Answer:

#5+20+80+...+20480# can be written as, #5\times1+5\times4^1+5\times4^2+...+5\times4^6#
#=5\times(1+4+4^2+...+4^6)=5\times(\frac{4^7-1}{4-1})=27305#.

Explanation:

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#.