lim _{x rightarrow 2}left(sum_{n=1}^{9} frac{ | vedclass

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Question

(A) Let $S = \lim _{x \rightarrow 2} \sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}$.Substitute $x = 2$ into the expression:$S = \sum_{n=1}^{9} \f

Vedclass · Accepted Answer

$\frac{9}{44}$

Vedclass · Answer

(A) Let $S = \lim _{x \rightarrow 2} \sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}$.Substitute $x = 2$ into the expression:$S = \sum_{n=1}^{9} \frac{2}{n(n+1)(4) + 2(2n+1)(2) + 4}$$S = \sum_{n=1}^{9} \frac{2}{4n^2 + 4n + 8n + 4 + 4} = \sum_{n=1}^{9} \frac{2}{4n^2 + 12n + 8}$$S = \sum_{n=1}^{9} \frac{2}{4(n^2 + 3n + 2)} = \sum_{n=1}^{9} \frac{1}{2(n+1)(n+2)}$Using partial fractions:$S = \frac{1}{2} \sum_{n=1}^{9} \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$This is a telescoping series:$S = \frac{1}{2} \left[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{10} - \frac{1}{11} \right) \right]$$S = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{11} \right) = \frac{1}{2} \left( \frac{11-2}{22} \right) = \frac{1}{2} \left( \frac{9}{22} \right) = \frac{9}{44}$.

$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :

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