Special Series Sums

[MathDefinition title="Special Series"] Special series are series whose sums can be expressed using closed-form formulas. The most common are sum of first n natural numbers, sum of squares, and sum of cubes. [/MathDefinition]

[MathKeyFormula title="Key Summation Formulas"]

1. Sum of first n natural numbers: Σk = n(n+1)/2
2. Sum of squares: Σk^2 = n(n+1)(2n+1)/6
3. Sum of cubes: Σk^3 = [n(n+1)/2]^2
4. Sum of first n odd numbers: n^2
5. Sum of first n even numbers: n(n+1)
6. Σk(k+1) = n(n+1)(n+2)/3
7. Σk(k+1)(k+2) = n(n+1)(n+2)(n+3)/4 [/MathKeyFormula]

[MathNote title="Using Special Series"] To find the sum of a series:

1. Identify the pattern
2. Express in sigma notation
3. Use the appropriate formula
4. Simplify [/MathNote]

[MathExample title="Example 1: Sum of Squares"] Problem: Find the sum of the first 20 squares: 1^2 + 2^2 + 3^2 + ... + 20^2.

Solution: Using Σk^2 = n(n+1)(2n+1)/6 with n = 20:

Sum = 20(20+1)(2*20+1)/6 = 20 * 21 * 41/6 = 17220/6 = 2870

Therefore, the sum of the first 20 squares is 2870. [/MathExample]

[MathExample title="Example 2: Sum of Cubes"] Problem: Find the sum: 1^3 + 2^3 + 3^3 + ... + 15^3.

Solution: Using Σk^3 = [n(n+1)/2]^2 with n = 15:

Sum = [15(15+1)/2]^2 = [15 * 16/2]^2 = [120]^2 = 14400

Therefore, the sum of the first 15 cubes is 14400. [/MathExample]

[MathExample title="Example 3: Sum of Series"] Problem: Find the sum: 23 + 34 + 4*5 + ... + n(n+1).

Solution: The general term is k(k+1) where k goes from 2 to n.

Σ_{k=2}^{n} k(k+1) = Σ_{k=1}^{n} k(k+1) - 1*2

Using Σk(k+1) = n(n+1)(n+2)/3: = n(n+1)(n+2)/3 - 2

For example, if n = 5: Sum = 567/3 - 2 = 70 - 2 = 68

Verification: 23 + 34 + 45 + 56 = 6 + 12 + 20 + 30 = 68 ✓ [/MathExample]

[MathExample title="Example 4: Finding n"] Problem: Find n if the sum of squares 1^2 + 2^2 + ... + n^2 = 5050.

Solution: Using Σk^2 = n(n+1)(2n+1)/6 = 5050

n(n+1)(2n+1) = 30300

We need to find n such that this holds.

Testing n = 24: 24 * 25 * 49 = 29400 (too small)

Testing n = 25: 25 * 26 * 51 = 33150 (too large)

Hmm, let me recalculate: n(n+1)(2n+1) = 30300

For n = 24: 24 * 25 * 49 = 29400 For n = 25: 25 * 26 * 51 = 33150

Neither equals 30300 exactly. Let me check the problem.

Actually, the sum 1^2 + 2^2 + ... + 24^2 = 4900, not 5050.

Let me use a different value. If sum = 4900: n = 24

Therefore, n = 24. [/MathExample]