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Sums with sigma notation practice problems

Evaluate sums in sigma notation with the formulas for Σi, Σi² and Σi³. 60 problems with worked solutions; in 60 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 4.1032

medium
Evaluate \( \displaystyle \sum_{i=2}^{25} \left(3 i^{2} + i + 5\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 2: compute from 1 to 25 and subtract the sum from 1 to 1.
    Reviewed
  3. \[ 5 \cdot 25 + 1 \cdot 25 \cdot 26 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 25 \cdot 26 \cdot 51 \cdot 1 \cdot \frac{1}{6} = 17025 \]
    From i = 1 to 25.✓ Proved
  4. \[ \sum_{i=1}^{1} \left(3 i^{2} + i + 5\right) = 9 \]
    From i = 1 to 1.✓ Proved
  5. \[ -9 + 17025 = 17016 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=2}^{25} \left(3 i^{2} + i + 5\right) = 17016 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 17016 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1034

medium
Evaluate \( \displaystyle \sum_{i=3}^{50} \left(- 2 i^{2} + 3 i - 4\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 50 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ \left(-2\right) 50 \cdot 51 \cdot 101 \cdot 1 \cdot \frac{1}{6} - 4 \cdot 50 + 3 \cdot 50 \cdot 51 \cdot 1 \cdot \frac{1}{2} = -82225 \]
    From i = 1 to 50.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(- 2 i^{2} + 3 i - 4\right) = -9 \]
    From i = 1 to 2.✓ Proved
  5. \[ -82225 + 9 = -82216 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{50} \left(- 2 i^{2} + 3 i - 4\right) = -82216 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -82216 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1036

medium
Evaluate \( \displaystyle \sum_{i=2}^{15} \left(i^{3} + 3 i^{2} + 4 i + 1\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 2: compute from 1 to 15 and subtract the sum from 1 to 1.
    Reviewed
  3. \[ 1 \cdot 15 + 4 \cdot 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} + \left(15 \cdot 16 \cdot 1 \cdot \frac{1}{2}\right)^{2} = 18615 \]
    From i = 1 to 15.✓ Proved
  4. \[ \sum_{i=1}^{1} \left(i^{3} + 3 i^{2} + 4 i + 1\right) = 9 \]
    From i = 1 to 1.✓ Proved
  5. \[ -9 + 18615 = 18606 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=2}^{15} \left(i^{3} + 3 i^{2} + 4 i + 1\right) = 18606 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 18606 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1039

medium
Evaluate \( \displaystyle \sum_{i=3}^{25} \left(i^{3} - i^{2} + 3 i + 2\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 25 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ \left(-1\right) 25 \cdot 26 \cdot 51 \cdot 1 \cdot \frac{1}{6} + 2 \cdot 25 + 3 \cdot 25 \cdot 26 \cdot 1 \cdot \frac{1}{2} + \left(25 \cdot 26 \cdot 1 \cdot \frac{1}{2}\right)^{2} = 101125 \]
    From i = 1 to 25.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(i^{3} - i^{2} + 3 i + 2\right) = 17 \]
    From i = 1 to 2.✓ Proved
  5. \[ -17 + 101125 = 101108 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{25} \left(i^{3} - i^{2} + 3 i + 2\right) = 101108 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 101108 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1234

medium
Evaluate \( \displaystyle \sum_{i=2}^{30} \left(- i^{2} + 2 i - 4\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 2: compute from 1 to 30 and subtract the sum from 1 to 1.
    Reviewed
  3. \[ \left(-1\right) 30 \cdot 31 \cdot 61 \cdot 1 \cdot \frac{1}{6} - 4 \cdot 30 + 2 \cdot 30 \cdot 31 \cdot 1 \cdot \frac{1}{2} = -8645 \]
    From i = 1 to 30.✓ Proved
  4. \[ \sum_{i=1}^{1} \left(- i^{2} + 2 i - 4\right) = -3 \]
    From i = 1 to 1.✓ Proved
  5. \[ -8645 + 3 = -8642 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=2}^{30} \left(- i^{2} + 2 i - 4\right) = -8642 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -8642 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1237

medium
Evaluate \( \displaystyle \sum_{i=2}^{15} \left(4 - i^{2}\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 2: compute from 1 to 15 and subtract the sum from 1 to 1.
    Reviewed
  3. \[ \left(-1\right) 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} + 4 \cdot 15 = -1180 \]
    From i = 1 to 15.✓ Proved
  4. \[ \sum_{i=1}^{1} \left(4 - i^{2}\right) = 3 \]
    From i = 1 to 1.✓ Proved
  5. \[ -1180 - 3 = -1183 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=2}^{15} \left(4 - i^{2}\right) = -1183 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -1183 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1240

medium
Evaluate \( \displaystyle \sum_{i=3}^{25} \left(- 3 i^{2} - 1\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 25 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ \left(-3\right) 25 \cdot 26 \cdot 51 \cdot 1 \cdot \frac{1}{6} - 1 \cdot 25 = -16600 \]
    From i = 1 to 25.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(- 3 i^{2} - 1\right) = -17 \]
    From i = 1 to 2.✓ Proved
  5. \[ -16600 + 17 = -16583 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{25} \left(- 3 i^{2} - 1\right) = -16583 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -16583 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1300

medium
Evaluate \( \displaystyle \sum_{i=2}^{30} \left(3 i^{2} + 2 i - 4\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 2: compute from 1 to 30 and subtract the sum from 1 to 1.
    Reviewed
  3. \[ \left(-4\right) 30 + 2 \cdot 30 \cdot 31 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 30 \cdot 31 \cdot 61 \cdot 1 \cdot \frac{1}{6} = 29175 \]
    From i = 1 to 30.✓ Proved
  4. \[ \sum_{i=1}^{1} \left(3 i^{2} + 2 i - 4\right) = 1 \]
    From i = 1 to 1.✓ Proved
  5. \[ -1 + 29175 = 29174 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=2}^{30} \left(3 i^{2} + 2 i - 4\right) = 29174 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 29174 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1307

medium
Evaluate \( \displaystyle \sum_{i=3}^{15} \left(- i^{2} - 3 i + 2\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 15 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ \left(-1\right) 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} - 3 \cdot 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 2 \cdot 15 = -1570 \]
    From i = 1 to 15.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(- i^{2} - 3 i + 2\right) = -10 \]
    From i = 1 to 2.✓ Proved
  5. \[ -1570 + 10 = -1560 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{15} \left(- i^{2} - 3 i + 2\right) = -1560 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -1560 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1309

medium
Evaluate \( \displaystyle \sum_{i=3}^{12} \left(3 i^{2} - i + 6\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 12 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ \left(-1\right) 12 \cdot 13 \cdot 1 \cdot \frac{1}{2} + 6 \cdot 12 + 3 \cdot 12 \cdot 13 \cdot 25 \cdot 1 \cdot \frac{1}{6} = 1944 \]
    From i = 1 to 12.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(3 i^{2} - i + 6\right) = 24 \]
    From i = 1 to 2.✓ Proved
  5. \[ -24 + 1944 = 1920 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{12} \left(3 i^{2} - i + 6\right) = 1920 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 1920 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1470

medium
Evaluate \( \displaystyle \sum_{i=3}^{25} \left(i^{2} + 2\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 3: compute from 1 to 25 and subtract the sum from 1 to 2.
    Reviewed
  3. \[ 2 \cdot 25 + 1 \cdot 25 \cdot 26 \cdot 51 \cdot 1 \cdot \frac{1}{6} = 5575 \]
    From i = 1 to 25.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(i^{2} + 2\right) = 9 \]
    From i = 1 to 2.✓ Proved
  5. \[ -9 + 5575 = 5566 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{25} \left(i^{2} + 2\right) = 5566 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 5566 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1474

medium
Evaluate \( \displaystyle \sum_{i=5}^{12} \left(i^{3} + 2 i^{2} + 3\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. The sum starts at i = 5: compute from 1 to 12 and subtract the sum from 1 to 4.
    Reviewed
  3. \[ 3 \cdot 12 + 2 \cdot 12 \cdot 13 \cdot 25 \cdot 1 \cdot \frac{1}{6} + \left(12 \cdot 13 \cdot 1 \cdot \frac{1}{2}\right)^{2} = 7420 \]
    From i = 1 to 12.✓ Proved
  4. \[ \sum_{i=1}^{4} \left(i^{3} + 2 i^{2} + 3\right) = 172 \]
    From i = 1 to 4.✓ Proved
  5. \[ -172 + 7420 = 7248 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=5}^{12} \left(i^{3} + 2 i^{2} + 3\right) = 7248 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 7248 \)

✓ Nihil obstat Lines: 4 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1033

easy
Evaluate \( \displaystyle \sum_{i=1}^{10} \left(3 i^{2} - 3 i + 5\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-3\right) 10 \cdot 11 \cdot 1 \cdot \frac{1}{2} + 5 \cdot 10 + 3 \cdot 10 \cdot 11 \cdot 21 \cdot 1 \cdot \frac{1}{6} = 1040 \]
    From i = 1 to 10.✓ Proved
  3. \[ \sum_{i=1}^{10} \left(3 i^{2} - 3 i + 5\right) = 1040 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 1040 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1035

easy
Evaluate \( \displaystyle \sum_{i=1}^{15} \left(3 i^{2} + 3 i\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ 3 \cdot 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} = 4080 \]
    From i = 1 to 15.✓ Proved
  3. \[ \sum_{i=1}^{15} \left(3 i^{2} + 3 i\right) = 4080 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 4080 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1037

easy
Evaluate \( \displaystyle \sum_{i=1}^{15} \left(5 - i\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-1\right) 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 5 \cdot 15 = -45 \]
    From i = 1 to 15.✓ Proved
  3. \[ \sum_{i=1}^{15} \left(5 - i\right) = -45 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -45 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1038

easy
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(- 2 i^{2} - 4 i\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-2\right) 25 \cdot 26 \cdot 51 \cdot 1 \cdot \frac{1}{6} - 4 \cdot 25 \cdot 26 \cdot 1 \cdot \frac{1}{2} = -12350 \]
    From i = 1 to 25.✓ Proved
  3. \[ \sum_{i=1}^{25} \left(- 2 i^{2} - 4 i\right) = -12350 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -12350 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1041

easy
Evaluate \( \displaystyle \sum_{i=1}^{10} \left(- i^{2} - 4 i - 2\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-1\right) 10 \cdot 11 \cdot 21 \cdot 1 \cdot \frac{1}{6} - 4 \cdot 10 \cdot 11 \cdot 1 \cdot \frac{1}{2} - 2 \cdot 10 = -625 \]
    From i = 1 to 10.✓ Proved
  3. \[ \sum_{i=1}^{10} \left(- i^{2} - 4 i - 2\right) = -625 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -625 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1043

easy
Evaluate \( \displaystyle \sum_{i=1}^{10} \left(3 i^{2} + 2 i + 6\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ 6 \cdot 10 + 2 \cdot 10 \cdot 11 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 10 \cdot 11 \cdot 21 \cdot 1 \cdot \frac{1}{6} = 1325 \]
    From i = 1 to 10.✓ Proved
  3. \[ \sum_{i=1}^{10} \left(3 i^{2} + 2 i + 6\right) = 1325 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 1325 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1044

easy
Evaluate \( \displaystyle \sum_{i=1}^{15} \left(- i^{2} - i + 3\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-1\right) 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} - 1 \cdot 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 3 \cdot 15 = -1315 \]
    From i = 1 to 15.✓ Proved
  3. \[ \sum_{i=1}^{15} \left(- i^{2} - i + 3\right) = -1315 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( -1315 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 4.1045

easy
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(5 i - 2\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
    Reviewed
  2. \[ \left(-2\right) 30 + 5 \cdot 30 \cdot 31 \cdot 1 \cdot \frac{1}{2} = 2265 \]
    From i = 1 to 30.✓ Proved
  3. \[ \sum_{i=1}^{30} \left(5 i - 2\right) = 2265 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 2265 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More sums with sigma notation practice

Evaluate \( \displaystyle \sum_{i=1}^{30} \left(- i^{2} - i\right) \) using summation formulas.
Problem 4.1040easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{12} \left(- 3 i^{2} + 4 i + 5\right) \) using summation formulas.
Problem 4.1046easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(i^{2} - 5 i + 6\right) \) using summation formulas.
Problem 4.1152easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{12} \left(3 i^{2} + 3 i\right) \) using summation formulas.
Problem 4.1153easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(3 i^{2} + 3 i + 3\right) \) using summation formulas.
Problem 4.1154easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(- 5 i\right) \) using summation formulas.
Problem 4.1155easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(i^{3} - 5 i\right) \) using summation formulas.
Problem 4.1156easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(- 3 i^{2} - 2 i - 2\right) \) using summation formulas.
Problem 4.1159easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(3 i^{2} + 2 i + 6\right) \) using summation formulas.
Problem 4.1160easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(i^{2} - 3 i - 4\right) \) using summation formulas.
Problem 4.1162easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{12} \left(4 - 3 i^{2}\right) \) using summation formulas.
Problem 4.1164easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(4 - 2 i\right) \) using summation formulas.
Problem 4.1165easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{12} \left(2 i - 2\right) \) using summation formulas.
Problem 4.1166easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(i^{3} + i^{2} - 5 i - 2\right) \) using summation formulas.
Problem 4.1231easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(- 2 i^{2} - 4 i - 2\right) \) using summation formulas.
Problem 4.1232easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(2 i^{2} + 4 i + 3\right) \) using summation formulas.
Problem 4.1236easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(3 i^{2} - 4 i + 1\right) \) using summation formulas.
Problem 4.1238easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(- 2 i^{2} - 4 i + 5\right) \) using summation formulas.
Problem 4.1239easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{10} \left(- i^{2} + 2 i - 1\right) \) using summation formulas.
Problem 4.1301easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(- 2 i^{2} + 2 i + 4\right) \) using summation formulas.
Problem 4.1302easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(- 3 i^{2} + 4 i + 5\right) \) using summation formulas.
Problem 4.1303easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(3 i^{2} + 5 i + 3\right) \) using summation formulas.
Problem 4.1304easy✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=1}^{50} \left(- i^{2} - i - 2\right) \) using summation formulas.
Problem 4.1306easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(- 2 i^{2} - 5 i - 2\right) \) using summation formulas.
Problem 4.1465easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(- i^{2} + 3 i + 4\right) \) using summation formulas.
Problem 4.1467easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{20} \left(3 i^{2} - 3 i - 2\right) \) using summation formulas.
Problem 4.1468easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{30} \left(- 3 i^{2} + 4 i + 3\right) \) using summation formulas.
Problem 4.1469easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{25} \left(- 3 i^{2} - i\right) \) using summation formulas.
Problem 4.1471easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{15} \left(- 5 i - 3\right) \) using summation formulas.
Problem 4.1472easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=1}^{10} \left(i^{3} - i^{2} - i\right) \) using summation formulas.
Problem 4.1473easy✓ Nihil obstat
Evaluate \( \displaystyle \sum_{i=5}^{15} \left(- 2 i^{2} + 4 i + 6\right) \) using summation formulas.
Problem 4.1042medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=3}^{15} \left(2 i^{2} - 4 i\right) \) using summation formulas.
Problem 4.1157medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=2}^{10} \left(- i\right) \) using summation formulas.
Problem 4.1158medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=3}^{30} \left(3 i^{2} - 3 i - 3\right) \) using summation formulas.
Problem 4.1161medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=5}^{12} \left(i^{3} - 3 i^{2} + 2\right) \) using summation formulas.
Problem 4.1163medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=2}^{15} \left(3 i^{2} - 3 i - 2\right) \) using summation formulas.
Problem 4.1233medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=2}^{15} \left(- 2 i^{2} - 4 i + 6\right) \) using summation formulas.
Problem 4.1235medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=2}^{20} \left(- 3 i^{2} + 5 i + 5\right) \) using summation formulas.
Problem 4.1305medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=5}^{30} \left(3 i^{2} + 3 i + 4\right) \) using summation formulas.
Problem 4.1308medium✓ Every equation proved
Evaluate \( \displaystyle \sum_{i=2}^{30} \left(i^{3} - 3 i^{2} - 2\right) \) using summation formulas.
Problem 4.1466medium✓ Every equation proved