Partial Sums of Arithmetic & Geometric Series

The fourth partial sum means we add the first four terms. The first four terms are 2, 5, 8, and 11. Adding them gives 2 + 5 = 7, 7 + 8 = 15, and 15 + 11 = 26. Therefore, S₄ = 26.

The formula for the nth partial sum of a geometric series is Sₙ = a₁(1 - rⁿ)/(1 - r). Here, a₁ = 3, r = 2, and n = 3. First, calculate rⁿ = 2³ = 8. Then, S₃ = 3(1 - 8)/(1 - 2) = 3(-7)/(-1) = 3(7) = 21. We can verify by listing terms: first term = 3, second = 3*2 = 6, third = 6*2 = 12. Sum = 3 + 6 + 12 = 21.

This is a geometric series with a₁ = 1/2 and r = 1/2. The formula for the nth partial sum is Sₙ = a₁(1 - rⁿ)/(1 - r). For n=5: S₅ = (½)(1 - (½)⁵) / (1 - 1/2) = (½)(1 - 1/32) / (½). Note that (½)/(½)=1, so S₅ = 1 - 1/32 = 31/32. We can also add the terms: 1/2=16/32, 1/4=8/32, 1/8=4/32, 1/16=2/32, 1/32=1/32. Sum = (16+8+4+2+1)/32 = 31/32.

For a convergent alternating series whose terms decrease in magnitude and approach zero, the Alternating Series Error Bound states that the error when using the nth partial sum to approximate the total sum is at most the absolute value of the next term. Here, the next term after the 10th partial sum is the 11th term, which has an absolute value of 0.021. Therefore, the maximum possible difference (error) is 0.021.

Recall that Sₙ is the sum of the first n terms. So S₁ = a₁ = 3. S₂ = a₁ + a₂ = 5, so a₂ = S₂ - a₁ = 5 - 3 = 2. S₃ = a₁ + a₂ + a₃ = 6, so a₃ = S₃ - S₂ = 6 - 5 = 1. Therefore, the third term is 1.

The alternating series error bound theorem states that when approximating the sum of a convergent alternating series with its nth partial sum, the maximum error is at most the absolute value of the first omitted term. For the fifth partial sum S5, we sum the first five terms (n=1 to n=5). The first omitted term is the sixth term (n=6), which equals (-½)5 = -1/32. The absolute value of this term is 1/32, so the maximum error when using S5 to approximate the sum is 1/32.

First, we find the first three terms. The term aₙ = 1/(n(n+1)) which can be rewritten using partial fractions as 1/n - 1/(n+1). So a₁ = 1/1 - 1/2 = 1/2. a₂ = 1/2 - 1/3 = 1/6. a₃ = 1/3 - 1/4 = 1/12. Now, S₃ = a₁ + a₂ + a₃ = 1/2 + 1/6 + 1/12. Convert to common denominator 12: 6/12 + 2/12 + 1/12 = 9/12 = 3/4.

For an arithmetic series, the sum of the first n terms is Sₙ = (n/2)(2a₁ + (n-1)d). Here, n=5, a₁=7, d=4. So S₅ = (5/2)(2*7 + (5-1)*4) = (5/2)(14 + 4*4) = (5/2)(14 + 16) = (5/2)(30) = 5 * 15 = 75. We can verify by listing terms: 7, 11, 15, 19, 23. Sum = 7+11=18, 18+15=33, 33+19=52, 52+23=75.

By definition, the nth partial sum Sₙ is the sum of the first n terms of a series. Option A is false because many series diverge (e.g., 1+1+1+...). Option B is true only for the sum of the infinite series, not the partial sum formula, but the statement says "partial sums... converge to a₁/(1-r)" which is the limit, so it's about the limit of partial sums, but the wording is tricky. However, the most directly and always true statement is C. Option D is false (harmonic series is a counterexample).

The notation S_20 represents the sum of the first 20 terms (a₁ + a_2 + ... + a_20). The notation S_9 represents the sum of the first 9 terms (a₁ + a_2 + ... + a_9). To isolate the sum of terms starting from a₁0 and ending at a_20, we must take the total sum S_20 and subtract the terms that come before a₁0. The terms before a₁0 are exactly the first 9 terms, represented by S_9. Therefore, the correct expression is S_20 - S_9.

The formula Sₙ = n(n+1)/2 is the formula for the sum of the first n positive integers. The series 1+2+3+4+... is exactly that. For option B, the sum of the first n odd numbers is n². For option C, the sum of the first n squares is n(n+1)(2n+1)/6. For option D, the harmonic series has no simple closed form for partial sums.

The partial sum that includes terms up to n=3 means we sum the terms for n=0,1,2,3. The next term is for n=4. The absolute value of the term for n=4 is |(-1)⁴ / 4!| = 1/24. By the alternating series error bound, the maximum error is the absolute value of the first omitted term, which is 1/24.

The value of the infinite series is the limit of the partial sums as k approaches infinity. So we compute limk→∞ Sₖ = limk→∞ (3k/(k+1)). Divide numerator and denominator by k: = limk→∞ (3/(1+1/k)) = 3/(1+0) = 3. So the series converges to 3.

The error bound from the integral test for the series Σ 1/n² is Rₙ ≤ ∫ₙ∞ (1/x²) dx. The antiderivative of 1/x² is -1/x, so the improper integral evaluates to [-1/x]N∞, which is 0 - (-1/N) = 1/N. We require the error to be less than 0.01, so we must solve the inequality 1/N < 0.01. This implies N > 1/0.01, or N > 100. We must choose a number of terms N that is strictly greater than 100. From the given options, N=10 and N=100 are not sufficient. N=1000 is the smallest option that satisfies the condition N > 100.

We write out the first few terms of the partial sum to identify the pattern. The first term is (1/1 - 1/2). The second term is (1/2 -⅓). The third term is (1/3 - 1/4), and the n-th term is (1/n - 1/(n+1)). When we add these up for Sₙ, the -1/2 from the first term cancels with the positive 1/2 from the second term. The -1/3 from the second term cancels with the positive 1/3 from the third term. This cancellation continues until only the very first part of the first term, which is 1, and the very last part of the n-th term, which is -1/(n+1), remain. Therefore, the formula simplifies to 1 - 1/(n+1).