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COMEDK · Maths · 5. Sequences and Series

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

  1. A \(\dfrac{120}{11}\)
  2. B \(\dfrac{13}{11}\)
  3. C \(\dfrac{110}{11}\)
  4. D \(\dfrac{19}{11}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\dfrac{120}{11}\)

Step-by-step Solution

Detailed explanation

The general term \(T_n\) of the series is given by \(T_n = \dfrac{(n+1)^2 - n}{n(n+1)}\). Simplifying the numerator: \((n+1)^2 - n = n^2 + 2n + 1 - n = n^2 + n + 1\). Thus, \(T_n = \dfrac{n^2 + n + 1}{n(n+1)} = \dfrac{n(n+1) + 1}{n(n+1)} = 1 + \dfrac{1}{n(n+1)}\). Using partial…