(1 + 2i2 + 4i4 + 8i6+.....)  का मान क्या है, जहाँ i = \(\sqrt {-1}\) है?

  1. \(\rm \frac i2\)
  2. \(\rm \frac1 3\)
  3. \(\rm \frac 4i\)
  4. इनमें से कोई नहीं 

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac1 3\)
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Detailed Solution

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संकल्पना:

ज्यामितीय श्रेणी में अनंत पदों की संख्या को निम्न द्वारा ज्ञात किया गया है, \(\rm S_\infty= \frac{a}{1-r}\), जहाँ, a = पहला पद और r = सार्व अनुपात

i = \(\sqrt {-1}\), i2 = -1

गणना:

दी गयी श्रृंखला (1 + 2i2 + 4i4 + 8i6+.....) है। 

स्पष्ट रूप से यह a = 1 और r = 2i2 के साथ एक ज्यामितीय श्रेणी है। 

अब योग = \(\rm S_\infty= \frac{a}{1-r}\)

 \(\rm = \frac{1}{1-2i^2} \\=\frac{1}{1+2}\)

\(\rm \frac1 3\)

∴ (1+ 2i2 + 4i4 + 8i6+.....) = 1/3

अतः विकल्प (2) सही है। 

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