ધન પૂર્ણાંકો n માટે,જો 4 a_n = (n^2 + 5n + 6) | vedclass

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(C) $a_n = \frac{n^2 + 5n + 6}{4} = \frac{(n+2)(n+3)}{4}$ $S_n = \sum_{k=1}^n \frac{4}{(k+2)(k+3)}$ $S_n = 4 \sum_{k=1}^n \left( \frac{1}{k+2} - \frac

Vedclass · Accepted Answer

$675$

Vedclass · Answer

(C) $a_n = \frac{n^2 + 5n + 6}{4} = \frac{(n+2)(n+3)}{4}$ $S_n = \sum_{k=1}^n \frac{4}{(k+2)(k+3)}$ $S_n = 4 \sum_{k=1}^n \left( \frac{1}{k+2} - \frac{1}{k+3} ight)$ તફાવતની રીતનો ઉપયોગ કરતા: $S_n = 4 \left[ \left( \frac{1}{3} - \frac{1}{4} ight) + \left( \frac{1}{4} - \frac{1}{5} ight) + \dots + \left( \frac{1}{n+2} - \frac{1}{n+3} ight) ight]$ $S_n = 4 \left( \frac{1}{3} - \frac{1}{n+3} ight) = 4 \left( \frac{n+3-3}{3(n+3)} ight) = \frac{4n}{3(n+3)}$ $n = 2025$ માટે: $S_{2025} = \frac{4 	imes 2025}{3(2025+3)} = \frac{4 	imes 2025}{3 	imes 2028} = \frac{4 	imes 2025}{6084}$ $507 S_{2025} = 507 	imes \frac{4 	imes 2025}{3 	imes 2028} = 675$

ધન પૂર્ણાંકો $n$ માટે,જો $4 a_n = (n^2 + 5n + 6)$ અને $S_n = \sum_{k=1}^n \left(\frac{1}{a_k}\right)$ હોય,તો $507 S_{2025}$ નું મૂલ્ય કેટલું થાય?

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