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Essential Summation Formulas in Calculus

Mar 15, 2025

Summation Formulas in Calculus

Introduction

  • Focus on summation formulas essential for calculus, especially for calculating the area of a region.
  • Evaluation of sums using sigma notation.

Summation of a Constant

  • Formula: If (i) runs from 1 to (n) for a constant (c), then the sum equals (c \times n).
    • Example: Sum of 8 over 4 terms is (8 \times 4 = 32).
    • Example Problems:
      • (c = 7), (n = 8): (7 \times 8 = 56).
      • (c = 9), (n = 5): (9 \times 5 = 45).

Summation of Integers

  • Formula: Sum of integers (i) = (\frac{n(n+1)}{2}).
    • Example: Sum from 1 to 5 is (\frac{5(5+1)}{2} = 15).
    • Confirmation with actual addition: 1 + 2 + 3 + 4 + 5 = 15.

Sum of Multiples

  • Example: Sum from 1 to 4 of 6(i) terms.
    • Calculation: 6 + 12 + 18 + 24 = 60.
    • Using formula: (6 \times \frac{4(4+1)}{2} = 60).

Sum of Linear Expressions

  • Example: Sum from 1 to 5 of (7i - 3).
    • Sequence: 4, 11, 18, 25, 32.
    • Sum: 90.
    • Using formulas:
      • Break into parts: (7 \times \frac{5(5+1)}{2} - 3 \times 5).
      • Result: 105 - 15 = 90.

Practicing with Adjustments

  • Example: Sum from 1 to 8 of (9i + 7).
    • Calculated: 380.
    • Using formula: Confirmed via actual adding each term.

Summation of Squares

  • Formula: Sum of squares (i^2 = \frac{n(n+1)(2n+1)}{6}).
    • Example: Sum of squares to 6:
      • Calculation: 1, 4, 9, 16, 25, 36 = 91.
      • Formula: (\frac{6(6+1)(2 \times 6+1)}{6} = 91).

Summation of Cubes

  • Formula: Sum of cubes (i^3 = \frac{n^2(n+1)^2}{4}).
    • Example: Sum of cubes to 4:
      • 1, 8, 27, 64 = 100.
      • Formula: ((4^2 \times 5^2)/4 = 100).

Practice Examples

  • Example: Sum of (i(i^2 + 4i)).
    • Distribute and separate: (i^3 + 4i^2).
    • Evaluate each part:
      • (\frac{n^2(n+1)^2}{4} + 4\frac{n(n+1)(2n+1)}{6}).
      • Result for (n=5): 445.

Final Practice

  • Example: Sum from 1 to 9 of ((i-1)^2).
    • Expand and separate:
      • (i^2 - 2i + 1).
    • Evaluate using formulas for each term.
    • Result: 204.

Conclusion

  • Understanding and applying different summation formulas can simplify evaluations in calculus, especially for identifying areas under curves and other advanced calculations.