Question
Download Solution PDF(1 + 2i2 + 4i4 + 8i6+.....) का मान क्या है, जहाँ i = \(\sqrt {-1}\) है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
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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