If Z=12√+i2√𝑍=12+𝑖2, then the value of ∣∣1+Z2+Z4+Z6+Z8+....+Z2008∣∣1+𝑍2+𝑍4+𝑍6+𝑍8+....+𝑍2008 is equal to
Question
If Z=12√+i2√Z=12+i2, then the value of ∣∣1+Z2+Z4+Z6+Z8+....+Z2008∣∣1+Z2+Z4+Z6+Z8+....+Z2008 is equal to
Solution
The given expression is a geometric series with first term as 1 and common ratio as Z^2.
The sum of an infinite geometric series with first term 'a' and common ratio 'r' (where |r| < 1) is given by a / (1 - r).
Here, a = 1 and r = Z^2 = (12 + i2)^2 = 144 - 4 + 24i = 140 + 24i.
The magnitude of r = |140 + 24i| = √(140^2 + 24^2) = √(19600 + 576) = √20276 = 142.4.
Since |r| > 1, the given series does not converge. Therefore, the sum of the series cannot be determined.
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