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
- A \(\dfrac{120}{11}\)
- B \(\dfrac{13}{11}\)
- C \(\dfrac{110}{11}\)
- D \(\dfrac{19}{11}\)
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…
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