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(1+ 2i2 + 4i4 + 8i6+.....) = Where i = \(\sqrt {-1}\)1. \(\rm \frac i2\)2. \(\rm \frac1 3\)3. \(\rm \frac 4i\)4. None of these |
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Answer» Correct Answer - Option 2 : \(\rm \frac1 3\) Concept: Sum of infinite terms in geometric progression is given by, \(\rm S_\infty= \frac{a}{1-r}\), where, a = first term and r = common ratio i = \(\sqrt {-1}\), i2 = -1 Calculation: Given series is (1 + 2i2 + 4i4 + 8i6+.....) Clearly it is a GP, with a = 1 and r = 2i2 Now sum = \(\rm S_\infty= \frac{a}{1-r}\) \(\rm = \frac{1}{1-2i^2} \\=\frac{1}{1+2}\) = \(\rm \frac1 3\) Hence, option (2) is correct. |
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