\(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16}+\)_____ के पहले 4 पदों का योग है:

  1. \(\dfrac{5}{164}\)
  2. \(\dfrac{39}{41}\)
  3. \(\dfrac{39}{320}\)
  4. \(\dfrac{1}{20}\)

Answer (Detailed Solution Below)

Option 4 : \(\dfrac{1}{20}\)
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गणना:

\(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16}\)

\(\dfrac{1}{4\times 8} + {(\dfrac{1}{1} + \dfrac{1}{3}+\dfrac{1}{6} +\dfrac{1}{10} )}\)

\(\dfrac{1}{4\times 8} {(\dfrac{1}{1} + \dfrac{1}{3 \times 1}+\dfrac{1}{3 \times 2}+\dfrac{1}{2 \times 5})}\)

\(\dfrac{1}{4\times 8} {(\dfrac{96}{60})} = 1/20\)

 

Alternate Method

\(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16} + ...\)

प्रत्येक पद  \(1 \over {4n \times 4(n + 1)}\) स्वरूप में है

⇒ चौथे पद n के लिए = 4

⇒ चौथा पद होना चाहिए \({1 \over {(4 \times 4) \times [4 \times (4 + 1)}]}= {1 \over {16 \times 20}}\)

श्रृंखला हो जाएगी \(\dfrac{1}{4\times 8} + \dfrac{1}{8\times 12} + \dfrac{1}{12\times 16} + \dfrac{1}{16\times 20}\)

⇒ \({1 \over 4}[({1 \over 4} - {1 \over 8}) + ({1 \over 8} - {1 \over 12}) + ({1 \over 12} - {1 \over 16}) + ({1 \over 16} - {1 \over 20})]\)

⇒ \({1 \over 4} \times ({1 \over 4} - {1 \over 20})\)

⇒ \({1 \over 4} \times {4 \over 20}\)

⇒ 1/20

श्रृंखला के पहले चार पदों का अभीष्ट योग 1/20 है

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