Q94 / 180 · COMEDK 2022 · Mathematics
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
- Correct answerA$\dfrac{120}{11}$
- B$\dfrac{13}{11}$
- C$\dfrac{110}{11}$
- D$\dfrac{19}{11}$
Solution
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 fractions, $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$.
Therefore, $T_n = 1 + \dfrac{1}{n} - \dfrac{1}{n+1}$.
The sum of 10 terms is $S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} (1 + \dfrac{1}{n} - \dfrac{1}{n+1})$.
$S = \sum_{n=1}^{10} 1 + \sum_{n=1}^{10} (\dfrac{1}{n} - \dfrac{1}{n+1})$.
The first part is $10 \times 1 = 10$.
The second part is a telescoping sum: $(1 - \dfrac{1}{2}) + (\dfrac{1}{2} - \dfrac{1}{3}) + \ldots + (\dfrac{1}{10} - \dfrac{1}{11}) = 1 - \dfrac{1}{11} = \dfrac{10}{11}$.
$S = 10 + \dfrac{10}{11} = \dfrac{110 + 10}{11} = \dfrac{120}{11}$.
Answer: $\dfrac{120}{11}$
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 fractions, $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$.
Therefore, $T_n = 1 + \dfrac{1}{n} - \dfrac{1}{n+1}$.
The sum of 10 terms is $S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} (1 + \dfrac{1}{n} - \dfrac{1}{n+1})$.
$S = \sum_{n=1}^{10} 1 + \sum_{n=1}^{10} (\dfrac{1}{n} - \dfrac{1}{n+1})$.
The first part is $10 \times 1 = 10$.
The second part is a telescoping sum: $(1 - \dfrac{1}{2}) + (\dfrac{1}{2} - \dfrac{1}{3}) + \ldots + (\dfrac{1}{10} - \dfrac{1}{11}) = 1 - \dfrac{1}{11} = \dfrac{10}{11}$.
$S = 10 + \dfrac{10}{11} = \dfrac{110 + 10}{11} = \dfrac{120}{11}$.
Answer: $\dfrac{120}{11}$
Hence, the correct answer is option: 'A'.