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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Limits

lim_{x \to 2} (\sum_{n = 1}^{9} \frac{x}{n ( n + 1 ) x ^{2} + 2 ( 2 n + 1 ) x + 4})

is equal to:

(A) $\frac{9}{44}$ ✓ Correct
(B) $\frac{5}{24}$
(C) $\frac{7}{36}$
(D) $\frac{1}{5}$

Answer: (A)

Solution

Substitute

x = 2

directly:

\sum_{n = 1}^{9} \frac{2}{n ( n + 1 ) \cdot 4 + 2 ( 2 n + 1 ) \cdot 2 + 4} = \sum_{n = 1}^{9} \frac{2}{4 n ( n + 1 ) + 4 ( 2 n + 1 ) + 4} = \sum_{n = 1}^{9} \frac{2}{4 n ^{2} + 4 n + 8 n + 4 + 4} = \sum_{n = 1}^{9} \frac{2}{4 n ^{2} + 12 n + 8} = \sum_{n = 1}^{9} \frac{1}{2 n ^{2} + 6 n + 4} = \sum_{n = 1}^{9} \frac{1}{2 ( n + 1 ) ( n + 2 )}

. Using partial fractions:

\frac{1}{2 ( n + 1 ) ( n + 2 )} = \frac{1}{2} [\frac{1}{n + 1} - \frac{1}{n + 2}]

. This telescopes:

\frac{1}{2} [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{10} - \frac{1}{11})] = \frac{1}{2} (\frac{1}{2} - \frac{1}{11}) = \frac{1}{2} \cdot \frac{9}{22} = \frac{9}{44}

.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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