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(1-1/4)x(1-1/9)x(1-1/16)x(1-1/25)x(1-1/36)x(1-1/49)

3 min read Jun 16, 2024
(1-1/4)x(1-1/9)x(1-1/16)x(1-1/25)x(1-1/36)x(1-1/49)

Unraveling the Pattern: (1-1/4)x(1-1/9)x(1-1/16)x(1-1/25)x(1-1/36)x(1-1/49)

This seemingly complex multiplication problem hides a beautiful pattern that makes solving it surprisingly simple. Let's break it down step by step:

Recognizing the Pattern

First, notice that each term in the multiplication is of the form (1 - 1/n²), where 'n' is a consecutive integer starting from 2.

  • (1 - 1/4) = (1 - 1/2²)
  • (1 - 1/9) = (1 - 1/3²)
  • (1 - 1/16) = (1 - 1/4²)
  • (1 - 1/25) = (1 - 1/5²)
  • (1 - 1/36) = (1 - 1/6²)
  • (1 - 1/49) = (1 - 1/7²)

Simplifying the Terms

Now, let's simplify each term using the difference of squares factorization: a² - b² = (a + b)(a - b)

  • (1 - 1/2²) = (1 + 1/2)(1 - 1/2) = (3/2)(1/2)
  • (1 - 1/3²) = (1 + 1/3)(1 - 1/3) = (4/3)(2/3)
  • (1 - 1/4²) = (1 + 1/4)(1 - 1/4) = (5/4)(3/4)
  • (1 - 1/5²) = (1 + 1/5)(1 - 1/5) = (6/5)(4/5)
  • (1 - 1/6²) = (1 + 1/6)(1 - 1/6) = (7/6)(5/6)
  • (1 - 1/7²) = (1 + 1/7)(1 - 1/7) = (8/7)(6/7)

The Magic of Cancellation

Notice that in the resulting multiplication, a lot of terms cancel out:

(3/2)(1/2) x (4/3)(2/3) x (5/4)(3/4) x (6/5)(4/5) x (7/6)(5/6) x (8/7)(6/7)

  • The 2 in the denominator of the first term cancels with the 2 in the numerator of the second term.
  • The 3 in the denominator of the second term cancels with the 3 in the numerator of the third term.
  • This pattern continues throughout the entire multiplication.

The Final Answer

After all the cancellations, we are left with:

(8/7)

Therefore, (1-1/4)x(1-1/9)x(1-1/16)x(1-1/25)x(1-1/36)x(1-1/49) = 8/7

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