Question

Show that 9+90+900+...+9x19^n-1=10^n -1 (Arithmetic and geometric series and their sums)

Answer 1

what about 19^n-1, i think it should be 10^n-1
9+90+900+...n terms
=9(1+10+100+...n terms) ...(1)
\[1+10+100+... n terms.\]
it is a G.P.
first term a=1
common ratio r=10
\[S=a \frac{ r ^{n}-1 }{r-1 }\]
find S and put in (1)

Answer 2

let S = 9 + 90 + 900 + ...
S = 9(10)^0 + 9(10)^1 + 9(10)^2 + .. + 9(10)^(n-1)

then
10S = 9(10)^1 + 9(10)^2 + 9(10)^3 + .. + 9(10)^n

subtract the two equations:
10S - S = 9(10)^n - 9

9S = 9(10)^n - 9

S = 10^n - 1