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)
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?
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.
ohh, i thought 1-( -1/3)= 1+(1/3)
thanks so much
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...
It seems that you made a mistake in calculating the sum of the first 6 terms of the series. Let's break down the correct steps to find the difference between the sum of the infinite geometric series and the sum of the first 6 terms.
To find the sum of an infinite geometric series, we use the formula:
S = a / (1 - r)
where 'a' is the first term and 'r' is the common ratio.
In this case, the first term 'a' is 18 and the common ratio 'r' is -1/3.
Using the formula, we can calculate:
S = 18 / (1 - (-1/3))
S = 18 / (1 + 1/3)
S = 18 / (4/3)
S = 18 * (3/4)
S = 54/4
S = 13.5
So, the sum of the infinite geometric series is 13.5.
Now, to find the sum of the first 6 terms of the series, we can sum them directly:
Sn = 18 - 6 + 2 - 2/3 + ...
Calculating the sum of the first 6 terms:
Sn = 18 - 6 + 2 - 2/3 + (-2/3)^2 + (-2/3)^3
Sn = 18 - 6 + 2 - 2/3 + (4/9) - (8/27)
Sn = 18 - 6 + 2 - 2/3 + 4/9 - 8/27
Sn = 20 - 2/3 + 4/9 - 8/27
Sn = (540 - 54 + 120 - 40) / 27
Sn = 566 / 27
Sn ≈ 20.963
The difference between the sum of the infinite series and the sum of the first 6 terms is:
13.5 - 20.963 ≈ -7.463
Therefore, the correct difference is approximately -7.463.