JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 16)
$$\mathop {\lim }\limits_{x \to 2} \left( {\sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} } \right)$$ is equal to :
$${9 \over {44}}$$
$${5 \over {24}}$$
$${1 \over 5}$$
$${7 \over {36}}$$
Explanation
$$S = \mathop {\lim }\limits_{x \to 2} \sum\limits_{n = 1}^9 {{x \over {n(n + 1){x^2} + 2(2n + 1)x + 4}}} $$
$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} $$
$$S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}$$
$$S = \sum\limits_{n = 1}^9 {{2 \over {4({n^2} + 3n + 2)}}} = {1 \over 2}\sum\limits_{n = 1}^9 {\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} $$
$$S = {1 \over 2}\left( {{1 \over 2} - {1 \over {11}}} \right) = {9 \over {44}}$$
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