determine the difference between the sum of the infinite geometric series 18-6+2-2/3...and the sum of the first 6 terms of the series (round to 3 decimal places)

Question

first i did
S=18/ (1+1/3)= 13 1/2
Sn= 18(1+(1/3)^6)
_____________
1 + 1/3

= 13 14/27

13 1/2 - 13 14/27 = -1/54

what did i do wrong?

Answer 1

your formula for Sn is wrong
it is

Sn = a(1 - r^n)/(1-r)
= 18(1 - (-1/3)^6)/(1+1/3)
= 18 (1 - 1/729)/(4/3)
= 18(728/729)(3/4)
= 364/27

((you had 365/27))

so the difference is │364/27 - 27/2│
= 1/54

It didn't say which way the subtraction was supposed to go, so I took the absolute value.

Answer 2

ohh, i thought 1-( -1/3)= 1+(1/3)

thanks so much

Answer 3

not 1-( -1/3) but 1-( -1/3)^6

you have to do the power first, so (-1/3)^6 = +1/729, now you subtract it...