Question
Download Solution PDFश्रृंखला 22 + 42 + 62 + .... + (2n)2 का योग ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
प्रथम n प्राकृत संख्याओं के वर्गों का योग निम्न द्वारा दिया जाता है:
\(\sum_{k=1}^{n}k^2=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\)
गणना:
माना Tn श्रृंखला का nवाँ पद है। तब,
Tn = (2n)2 = 4n2
∴ 22 + 42 + 62 + .... + (2n)2 = \(\sum_{k=1}^{n}T_{k}\)
\(\Rightarrow \sum_{k=1}^{n}4k^2\)
\(\Rightarrow 4\sum_{k=1}^{n}k^2\)
\(\Rightarrow 4\left \{ \frac{n(n+1)(2n+1)}{6} \right \}\)
\(\Rightarrow \frac{2}{3}\: n(n+1)(2n+1)\)
इसलिए, श्रृंखला 22 + 42 + 62 + .... + (2n)2 का योग
\(\frac{2}{3}\: n(n+1)(2n+1)\) है।
Last updated on May 31, 2025
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