To find the sum of this infinite series, notice that each term can be written as a fraction with a denominator of 5 raised to the power of the term number.
Using the formula for the sum of an infinite geometric series, we can find the sum:
S = a / (1 - r),
where S is the sum, a is the first term, and r is the ratio between consecutive terms.
In this case, a = 1/5 and r = 1/5^2 = 1/25.
S = (1/5) / (1 - 1/25),
S = (1/5) / (24/25),
S = (1/5) * (25/24),
S = 5/24.
Therefore, the sum of the series is 5/24.
find the sum of all the terms of
1/5 + 2/25 + 3/125 + 4/225 + .....
5 answers
I posted this question right after I gave you a similar type.
You totally botched the first solution, and I then gave you the correct solution.
I was under the impression that you learned from previous mistakes, however you didn't learn one bit
You botched this one as well, but in a different way.
You totally botched the first solution, and I then gave you the correct solution.
I was under the impression that you learned from previous mistakes, however you didn't learn one bit
You botched this one as well, but in a different way.
Apologies for the previous mistakes. Let's correct the solution for this series.
The series is given by 1/5 + 2/25 + 3/125 + 4/225 + ...
To find the sum of this series, we can consider it as the sum of two separate series:
Series A: 1/5 + 2/25 + 3/125 + 4/625 + ...
Series B: 0 + 0/25 + 0/125 + 0/625 + ...
Series B is a geometric series with a common ratio of 0, so it sums up to 0.
Now, let's find the sum of Series A. We can multiply each term by 1/5 to get a geometric series:
(1/5)(1/5) + (2/5)(1/25) + (3/5)(1/125) + (4/5)(1/625) + ...
This can be rewritten as:
1/25 + 2/125 + 3/625 + 4/3125 + ...
Now, we have a geometric series with a common ratio of 1/5. Using the formula for the sum of an infinite geometric series:
Sum of Series A = (a / (1 - r)) * (1/5),
where a is the first term and r is the ratio between consecutive terms.
In this case, a = 1/25 and r = 1/5.
Sum of Series A = ((1/25) / (1 - 1/5)) * (1/5),
Sum of Series A = ((1/25) / (4/5)) * (1/5),
Sum of Series A = (1/25) * (5/4) * (1/5),
Sum of Series A = 1/4.
Since Series B sums up to 0, the overall sum of the series is:
Sum = Sum of Series A + Sum of Series B,
Sum = 1/4 + 0,
Sum = 1/4.
Therefore, the sum of the given series is 1/4.
The series is given by 1/5 + 2/25 + 3/125 + 4/225 + ...
To find the sum of this series, we can consider it as the sum of two separate series:
Series A: 1/5 + 2/25 + 3/125 + 4/625 + ...
Series B: 0 + 0/25 + 0/125 + 0/625 + ...
Series B is a geometric series with a common ratio of 0, so it sums up to 0.
Now, let's find the sum of Series A. We can multiply each term by 1/5 to get a geometric series:
(1/5)(1/5) + (2/5)(1/25) + (3/5)(1/125) + (4/5)(1/625) + ...
This can be rewritten as:
1/25 + 2/125 + 3/625 + 4/3125 + ...
Now, we have a geometric series with a common ratio of 1/5. Using the formula for the sum of an infinite geometric series:
Sum of Series A = (a / (1 - r)) * (1/5),
where a is the first term and r is the ratio between consecutive terms.
In this case, a = 1/25 and r = 1/5.
Sum of Series A = ((1/25) / (1 - 1/5)) * (1/5),
Sum of Series A = ((1/25) / (4/5)) * (1/5),
Sum of Series A = (1/25) * (5/4) * (1/5),
Sum of Series A = 1/4.
Since Series B sums up to 0, the overall sum of the series is:
Sum = Sum of Series A + Sum of Series B,
Sum = 1/4 + 0,
Sum = 1/4.
Therefore, the sum of the given series is 1/4.
nope!
I will come back in a few weeks to see if this page has shown any improvement.
I used to be one of the regular tutors for many years, but this webpage has
deteriorated badly over the last 2 years, all the human tutors have left, and I was
actually banned from posting under my previous name.
so with only 4 or 5 views for a typical question, and the bot getting many of the math
questions wrong, I will stay away for a few more weeks, then take a peek.
RIP, Jishka
I will come back in a few weeks to see if this page has shown any improvement.
I used to be one of the regular tutors for many years, but this webpage has
deteriorated badly over the last 2 years, all the human tutors have left, and I was
actually banned from posting under my previous name.
so with only 4 or 5 views for a typical question, and the bot getting many of the math
questions wrong, I will stay away for a few more weeks, then take a peek.
RIP, Jishka
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