Use the sum of the first 10 terms to approximate the sum of the series ∑∞,n=1, 1/(3^n+4^n). Estimate the error.
To approximate the sum of the series ∑∞,n=1, 1/(3^n+4^n), we will start by finding the sum of the first 10 terms.
The general term of the series is given by aₙ = 1/(3ⁿ+4ⁿ).
We want to find Sₙ , the sum of the first n terms of the series, which can be calculated using the formula:
Sₙ = a₁ + a₂ + a₃ + ... + aₙ
For our series, we have:
Sₙ = 1/(3²+4²) + 1/(3³+4³) + 1/(3⁴+4⁴) + ... + 1/(3ⁿ+4ⁿ)
To calculate the sum of the first 10 terms, we substitute n = 10 into the formula:
S₁₀ = 1/(3²+4²) + 1/(3³+4³) + 1/(3⁴+4⁴) + ... + 1/(3¹⁰+4¹⁰)
Now, let's compute the value of S₁₀.
S₁₀ = 1/(9+16) + 1/(27+64) + 1/(81+256) + ... + 1/(59049+1048576)
S₁₀ ≈ 0.066
Now, let's estimate the error.
To estimate the error in the approximation of the sum of the series, we use the remainder formula:
Rₙ = S - Sₙ
where Rₙ is the error, S is the exact sum of the series, and Sₙ is the sum of the first n terms of the series.
Since we have an infinite series, we cannot calculate the exact sum S. However, we can approximate the error by finding the difference between the sum of the first 10 terms and the sum of a larger number of terms.
Let's calculate the error by finding the sum of the first 1000 terms (S₁₀₀₀) and subtracting the sum of the first 10 terms (S₁₀).
S₁₀₀₀ = 1/(3²+4²) + 1/(3³+4³) + 1/(3⁴+4⁴) + ... + 1/(3¹⁰⁰⁰+4¹⁰⁰⁰)
S₁₀₀₀ = [calculations here]
S₁₀₀₀ ≈ [result here]
Now we can calculate the error:
R₁₀₀₀ = S - S₁₀₀₀
R₁₀₀₀ ≈ S - [result here]
Please note that the error estimation is an approximation itself as we are not able to find the exact sum of an infinite series.
To approximate the sum of the series ∑∞,n=1, 1/(3^n+4^n) using the sum of the first 10 terms, we need to find the sum of the first 10 terms and calculate the error between this approximate value and the actual sum of the series.
Let's start by finding the sum of the first 10 terms:
S10 = 1/(3^1+4^1) + 1/(3^2+4^2) + 1/(3^3+4^3) + ... + 1/(3^10+4^10)
Now, let's calculate the value of S10 by substituting each n value from 1 to 10:
S10 = 1/(3^1+4^1) + 1/(3^2+4^2) + 1/(3^3+4^3) + ... + 1/(3^10+4^10)
≈ 0.14463
Next, let's calculate the actual sum of the series. Since we're dealing with an infinite series, we cannot directly find the exact sum. However, we can estimate it by calculating the limit of the series as n approaches infinity:
lim (n→∞) Σ 1/(3^n+4^n)
To estimate this limit, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where 'a' is the first term and 'r' is the common ratio. In this case, a = 1/(3^1+4^1) = 1/7 and r = (1/3)/(1/4) = 4/3.
So,
S ≈ (1/7) / (1 - 4/3)
= (1/7) / (-1/3)
= -3/7
Now, let's calculate the error between S10 and the actual sum (S):
Error = |S10 - S|
= |0.14463 - (-3/7)|
= |0.14463 + 3/7|
= 0.71263
Therefore, the estimated error is approximately 0.71263.