Math Magic Wonderland
Discover the enchanting world of series summation with our friendly math wizard! ✨🧙♂️
Sum of Series (Multiple Parts)
Find the sum of these magical series:
- 1 + 2 + 3 + ... + 60
- 3 + 6 + 9 + ... + 96
- 51 + 52 + 53 + ... + 92
- 1 + 4 + 9 + 16 + ... + 225
- 6² + 7² + 8² + ... + 21²
- 10³ + 11³ + 12³ + ... + 20³
- 1 + 3 + 5 + ... + 71
Magical Formulas
These are your magic spells for solving series problems!
:
Sum of numbers: S = n(n+1)/2
Sum of squares: S = n(n+1)(2n+1)/6
Sum of cubes: S = [n(n+1)/2]²
Sum of numbers: S = n(n+1)/2
Sum of squares: S = n(n+1)(2n+1)/6
Sum of cubes: S = [n(n+1)/2]²
(i) Sum = n(n+1)/2 = 60×61/2 = 1830That's a lot of numbers to add up the normal way!
(ii) 3(1+2+...+32) = 3×(32×33/2) = 1584See how we factored out the 3? Clever!
(iii) (Sum to 92) - (Sum to 50) = (92×93/2)-(50×51/2) = 4278-1275 = 3003Subtracting sums is often easier than adding many numbers!
(iv) Sum of squares (1² to 15²) = 15×16×31/6 = 1240Square numbers make beautiful patterns!
(v) (Sum of 1² to 21²) - (Sum of 1² to 5²) = (21×22×43/6)-(5×6×11/6) = 3311-55 = 3256The difference between two sums gives us exactly what we need!
(vi) (Sum of 1³ to 20³) - (Sum of 1³ to 9³) = (20×21/2)²-(9×10/2)² = 44100-2025 = 42075Cubes grow really fast, don't they?
(vii) Sum of odd numbers = n² = 36² = 1296Beautiful how the sum of odd numbers forms perfect squares! (since there are 36 odd numbers from 1 to 71)
Sum of Cubes from Sum
If 1 + 2 + 3 + ... + k = 325, then find 1³ + 2³ + 3³ + ... + k³.
Magical RelationshipThe sum of cubes is the square of the sum of numbers!:
Scubes = (Snumbers)²
It's like math poetry!
Scubes = (Snumbers)²
It's like math poetry!
We know that 1 + 2 + ... + k = k(k+1)/2 = 325
The sum of cubes is equal to the squareIsn't it amazing that cubes and squares are related this way? of this sum
Therefore, 1³ + 2³ + ... + k³ = (325)² = 105,625That's a big number from a simple relationship!
Sum from Cubes
If 1³ + 2³ + 3³ + ... + k³ = 44100 then find 1 + 2 + 3 + ... + k.
Reverse the magic!
Snumbers = √(Scubes)
The square root unlocks the secret!
Snumbers = √(Scubes)
The square root unlocks the secret!
We know that 1³ + 2³ + ... + k³ = [k(k+1)/2]² = 44100
Taking square root of both sides: k(k+1)/2 = √44100 = 210The square root reveals the simpler sum underneath!
Therefore, 1 + 2 + ... + k = 210
Find Number of Terms
How many terms of the series 1³ + 2³ + 3³ + ... should be taken to get the sum 14400?
Sum of cubes formula: S = [n(n+1)/2]²
We'll work backwards to find n!
We'll work backwards to find n!
Given: [n(n+1)/2]² = 14400
Take square root: n(n+1)/2 = √14400 = 120The square root simplifies our problem!
Multiply by 2: n(n+1) = 240
This is n² + n - 240 = 0
Solving the quadratic equation: n = 15Only the positive solution makes sense here! (since n must be positive)
Find n from Cubes
The sum of the cubes of the first n natural numbers is 2025, then find the value of n.
Same magic formula: S = [n(n+1)/2]² = 2025
Let's find n!
Let's find n!
Given: [n(n+1)/2]² = 2025
Take square root: n(n+1)/2 = √2025 = 4545 is a triangular number too!
Multiply by 2: n(n+1) = 90
This is n² + n - 90 = 0
Solving the quadratic equation: n = 9The 9th triangular number squared gives us 2025! (since n must be positive)
Colorful Area Calculation
Rekha has 15 square color papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these color papers?
Total area = Sum of squares from 10² to 24²
= (Sum of 1² to 24²) - (Sum of 1² to 9²)
Smart subtraction!
= (Sum of 1² to 24²) - (Sum of 1² to 9²)
Smart subtraction!
We need to find 10² + 11² + ... + 24²
This equals (1² + 2² + ... + 24²) - (1² + 2² + ... + 9²)
Using sum of squares formula: n(n+1)(2n+1)/6
For 1-24: 24×25×49/6 = 4900That's a lot of square centimeters!
For 1-9: 9×10×19/6 = 285We subtract the squares we don't need
Total area = 4900 - 285 = 4615 cm²That's enough to decorate quite a space!
Sum of Difference Series
Find the sum of the series (2³ − 1³) + (4³ − 3³) + (6³ − 5³) + ... to:
- n terms
- 8 terms
Difference of cubes: a³ - b³ = (a - b)(a² + ab + b²)
But we'll find a smarter pattern!
But we'll find a smarter pattern!
General term: (2k)³ - (2k-1)³ = [8k³] - [8k³ - 12k² + 6k - 1] = 12k² - 6k + 1
(i) Sum to n terms: Σ(12k² - 6k + 1) from k=1 to n
= 12Σk² - 6Σk + Σ1 = 12[n(n+1)(2n+1)/6] - 6[n(n+1)/2] + n
= 2n(n+1)(2n+1) - 3n(n+1) + n = n(n+1)(4n+2-3) + n = n(n+1)(4n-1) + n
= n(4n² + 3n - 1 + 1) = n(4n² + 3n) = n²(4n + 3)
(ii) For 8 terms: 8²(4×8 + 3) = 64 × 35 = 2240The pattern holds beautifully!