#13

    Sopaantyadvayamantyam

    "The ultimate and twice the penultimate."

    💡 Quick Tip

    1/(2×3) + 1/(3×4) + 1/(4×5) → telescopes to 1/2 - 1/5 = 3/10

    What It's Used For

    Working with telescoping fraction sums and specific algebraic series.

    How It Works

    For certain fraction problems, the answer relates to the last term and twice the second-to-last term. Telescoping sums collapse beautifully.

    This sutra handles fraction sums of the form 1/(n(n+1)). Each such fraction can be decomposed as 1/n − 1/(n+1) using partial fractions. When you sum a series of these, most terms cancel in a "telescoping" fashion, leaving only the first and last terms. For example: 1/(1×2) + 1/(2×3) + 1/(3×4) = (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) = 1/1 − 1/4 = 3/4. The sutra also applies to dividing by 5: since there are two 5s in every 10, to divide by 5 you can double and divide by 10. The name "ultimate and twice the penultimate" refers to the relationship between the last terms in such series.

    History & Origin

    Telescoping series were studied by Indian mathematicians of the Kerala School (14th–16th century), particularly by Madhava of Sangamagrama who used them to derive infinite series for π and trigonometric functions — predating European discoveries by over 200 years. The Kerala School's work on infinite series using partial fraction decomposition is one of the most remarkable achievements in the history of mathematics. This sutra captures the essence of their technique in a simple memorable formula. The word "Sopaantyadvayamantyam" literally describes the structural relationship between consecutive and terminal terms.

    Scientific & Mathematical Basis

    The mathematical foundation is partial fraction decomposition: 1/(n(n+1)) = 1/n − 1/(n+1). Summing from n=a to n=b gives a telescoping series where all intermediate terms cancel, yielding 1/a − 1/(b+1). This technique is fundamental in calculus for evaluating definite integrals of rational functions. In physics, telescoping series appear in quantum mechanics (energy level transitions) and in electrical engineering (ladder network analysis). The technique is also used in computational complexity theory for analyzing recursive algorithms.

    Real-World Use Cases

    • Rapidly summing series of unit fractions — common in competitive mathematics
    • Physics: computing total resistance in ladder networks of resistors
    • Financial mathematics: present value calculations for annuities
    • Computer science: analyzing runtime of recursive algorithms via telescoping recurrences
    • Quick division by 5 using the "double and divide by 10" shortcut

    Step-by-Step Method

    1. 1Identify the ultimate (last) and penultimate (second-to-last) terms
    2. 2Apply partial fractions: 1/(n(n+1)) = 1/n - 1/(n+1)
    3. 3Most middle terms cancel (telescope), leaving first minus last

    📝 Worked Example

    Example 1: 1/(2×3) + 1/(3×4) + 1/(4×5)
    1. Step 1:Use partial fractions: 1/(n(n+1)) = 1/n - 1/(n+1)
    2. Step 2:Telescoping: 1/2 - 1/5
    3. Step 3:Answer: 3/10
    Answer: 3/10

    🌟 Beginner Level Examples

    Example 1: 1/(1×2) + 1/(2×3)
    1. Step 1:= (1/1 − 1/2) + (1/2 − 1/3)
    2. Step 2:Telescope: 1 − 1/3
    3. Step 3:Answer: 2/3
    Answer: 2/3
    Example 2: 1/(1×2) + 1/(2×3) + 1/(3×4)
    1. Step 1:Telescope: 1 − 1/4
    2. Step 2:Answer: 3/4
    Answer: 3/4

    🟡 Intermediate Level Examples

    Example 1: 1/(3×4) + 1/(4×5) + 1/(5×6)
    1. Step 1:Telescope: 1/3 − 1/6
    2. Step 2:Answer: 1/6
    Answer: 1/6
    Example 2: 85 ÷ 5 using the sutra
    1. Step 1:Double 85 → 170
    2. Step 2:Divide by 10 → 17
    3. Step 3:"Twice the penultimate": 2×8=16, plus 1 (from the 5) = 17
    Answer: 17

    🔴 Advanced Level Examples

    Example 1: Sum: 1/(5×6) + 1/(6×7) + ... + 1/(19×20)
    1. Step 1:Telescope: 1/5 − 1/20
    2. Step 2:= 4/20 − 1/20 = 3/20
    Answer: 3/20
    Example 2: 665 ÷ 5
    1. Step 1:Double: 1330
    2. Step 2:Divide by 10: 133
    Answer: 133

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