Math Problem Statement

If the pattern in the first three equations is continued, what will be the 20th equation?

If the pattern in the first three equations is continued, what will be the 20th equation?

Solution

The given image shows three equations with an increasing number of terms:

  1. 1+2=31 + 2 = 3
  2. 4+5+6=7+84 + 5 + 6 = 7 + 8
  3. 9+10+11+12=13+14+159 + 10 + 11 + 12 = 13 + 14 + 15

To determine the 20th equation in this pattern, we first need to identify the rule governing the number of terms and how they are arranged.

Observing the Pattern:

  1. Number of terms on each side:

    • 1st equation: 2 terms on the left, 1 term on the right.
    • 2nd equation: 3 terms on the left, 2 terms on the right.
    • 3rd equation: 4 terms on the left, 3 terms on the right.

    The number of terms on the left side starts at 2 and increases by 1 in each successive equation. The number of terms on the right side starts at 1 and also increases by 1 in each successive equation.

  2. The terms used:

    • The first equation uses numbers from 1 to 3.
    • The second equation uses numbers from 4 to 8.
    • The third equation uses numbers from 9 to 15.

    The numbers used in each equation are consecutive, and each equation starts from the next number after the last number used in the previous equation.

  3. Pattern Generalization: For the nn-th equation:

    • The left side has n+1n+1 terms.
    • The right side has nn terms.
    • The equation uses consecutive numbers starting from where the (n1)(n-1)-th equation ended.

Finding the 20th Equation:

  1. Number of terms:

    • Left side: 20+1=2120 + 1 = 21 terms.
    • Right side: 2020 terms.
  2. Finding the starting number:

    • We notice that each equation uses a block of consecutive numbers. To find the starting number for the 20th equation, we need to calculate how many numbers have been used up to the 19th equation.

    The total number of terms up to the 19th equation is: (2+3+4++20)=k=220k(2 + 3 + 4 + \ldots + 20) = \sum_{k=2}^{20} k

    This is an arithmetic series where the first term a=2a = 2, the last term l=20l = 20, and the number of terms n=19n = 19.

    The sum of this series is: S=n2×(a+l)=192×(2+20)=192×22=209S = \frac{n}{2} \times (a + l) = \frac{19}{2} \times (2 + 20) = \frac{19}{2} \times 22 = 209

    So, 209 numbers have been used up to the 19th equation. Therefore, the starting number for the 20th equation is 209+1=210209 + 1 = 210.

  3. Constructing the 20th equation:

    The 20th equation will start at 210 and will have 21 terms on the left and 20 terms on the right.

    • Left side terms: 210+211+212++230210 + 211 + 212 + \ldots + 230 (21 terms)
    • Right side terms: 231+232+233++250231 + 232 + 233 + \ldots + 250 (20 terms)

20th Equation:

210+211+212++230=231+232+233++250210 + 211 + 212 + \ldots + 230 = 231 + 232 + 233 + \ldots + 250

Would you like further details, or do you have any questions?

Related Questions:

  1. What is the sum of the left and right sides of the 20th equation?
  2. How many total terms are there in the first 20 equations combined?
  3. What is the 10th equation in the pattern?
  4. Can you derive a general formula for the starting number of the nn-th equation?
  5. How can we verify the equality in each equation without performing full addition?

Tip:

When analyzing patterns, look for both numerical progressions and changes in structure to identify the underlying rules.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Patterns

Formulas

Sum of an arithmetic series: S = (n / 2) * (a + l)
General nth term pattern for equation

Theorems

Arithmetic Series Theorem

Suitable Grade Level

Grades 8-10