Question

Find the sum of the finite geometric series
1) 1+1/2+1/4+1/8+...+1/2^8
2)30+3+3/10+3/100+....+3/1,000,000
3)10,000+1,000+100+10+,,,,,,,+1/10,000
4)1-1/2+1/4-1/8+.......+1/2^8

Answer 1

USE THIS G.P EQUATION
$a+ar+ar^2+..+ar^{n-1}= \frac{a(r^n-1)}{r-1}$

if r<1
$a+ar+ar^2+..+ar^{n-1}= \frac{a(1-r^n)}{1-r}$

first one:

$1+ \frac{1}{2}+ \frac{1}{4}+..+\frac{1}{2^8}= \frac{1- \frac{1}{2^9}}{1- \frac{1}{2}}=511/(2^8)=1.99609375$

second one:

$30+3+ \frac{3}{10}+ \frac{3}{100}+..+\frac{3}{10^6}=3[10+1+ \frac{1}{10}+ \frac{1}{100}+..+\frac{1}{10^6}]$
$=3\frac{10[1- \frac{1}{10^{8}}]}{1- \frac{1}{10}}= \frac{30(10^8-1)}{9(10^7)}=33.333333$

third one:
$10,000+1,000+100+10+,,,,,,,+ \frac{1}{10,000}= \frac{10000(1- \frac{1}{10^9} )}{1-\frac{1}{10}} =11111.1111$

fourth one:
$1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+..+-\frac{1}{2^8}=1+[-\frac{1}{2}]+[-\frac{1}{2}]^2+[-\frac{1}{2}]^3+..+[-\frac{1}{2}]^8=$
$\frac{1-[-\frac{1}{2}]^9}{1-[-\frac{1}{2}]}=0.66796875$