Question

If \( S = \frac{2^2 - 1}{2} + \frac{3^2 - 2}{6} + \frac{4^2 - 3}{12} + \ldots\) \text{ upto 10 terms, then S is equal to}

• A. \( \frac{120}{11} \)
• B. \( \frac{13}{11} \)
• C. \( \frac{110}{11} \)
• D. \( \frac{19}{11} \)

Hint:

For unusual sequences, try expressing them as simplified general terms or compute first few terms.

Answer 1

The Correct Option is A

Let’s write general term:
\( T_n = \frac{n^2 - (n-1)}{n(n-1)} = \frac{n^2 - n + 1}{n(n-1)} \)
Multiply numerator and simplify if needed. But better strategy: compute first 10 terms and observe pattern.
Computing sum of terms up to 10:
Calculate: \[ \frac{2^2 - 1}{2} = \frac{3}{2}, \quad \frac{3^2 - 2}{6} = \frac{7}{6}, \quad \frac{4^2 - 3}{12} = \frac{13}{12}, \ldots \] Summing all these manually (or using a pre-derived formula): \[ S = \frac{120}{11} \]