Question

The value of \(\displaystyle\sum_{k=5}^{36}\frac{1}{k^2-k}\) is equal to

• A. \(\frac{7}{36}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{2}{9}\)
• D. \(\frac{1}{12}\)
• E. \(\frac{5}{36}\)

Answer 1

The Correct Option is C

We are asked to find the value of the sum:

\(\sum_{k=5}^{36} \frac{1}{k^2 - k}\) 

First, factor the denominator:

\(k^2 - k = k(k - 1)\)

So, the general term in the sum is:

\(\frac{1}{k(k-1)}\)

Now, apply partial fraction decomposition to \(\frac{1}{k(k-1)}\):

\(\frac{1}{k(k-1)} = \frac{A}{k-1} + \frac{B}{k}\)

Multiplying both sides by \(k(k-1)\) gives:

\(1 = A \cdot k + B \cdot (k - 1)\)

Expanding and solving for \(A \) and \( B\):

\(1 = A \cdot k + B \cdot k - B\)

\(1 = (A + B)k - B\)

Equating coefficients, we get:

\(A + B = 0 \) and \( -B = 1\)

\(B = -1 \) and \( A = 1\)

Thus, we have:

\(\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}\)

Now, we can rewrite the sum as a telescoping series:

\(\sum_{k=5}^{36} \left( \frac{1}{k-1} - \frac{1}{k} \right)\)

Writing out the terms, we see most of the terms cancel, leaving us with:

\(\left( \frac{1}{4} - \frac{1}{36} \right)\)

Now, subtract:

\(\frac{1}{4} - \frac{1}{36} = \frac{9}{36} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9}\)

The answer is \( \frac{2}{9} \).