Question
Download Solution PDFश्रृंखला \(\rm \sqrt{2}+ \sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\) के पहले 24 पदों का योग क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसही उत्तर \(300\sqrt{2}\) है
संकल्पना:
1 से n तक क्रमागत संख्याओं का योग:\( \rm 1 + 2 + 3 +\ ...\ + n = \dfrac{n(n+1)}{2}\)
गणना:
दी गई श्रृंखला के पहले 24 पदों का योग इस प्रकार लिखा जा सकता है:
\(\rm \sqrt{2}+ \sqrt{8}+\sqrt{18}+\sqrt{32}+\ ...\)
\(\rm =\sqrt{2}+ 2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+\ ...\ +24\sqrt2\)
\(\rm =\sqrt2(1+2+\ ...\ +24)\)
\(\rm =\sqrt2\times \dfrac{24\times25}{2}=300\sqrt2\)
Last updated on Jul 4, 2025
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