SOLUTION TO BRAINTEASER NO. 3 DISCLAIMER This solution is but one of many possible methods for solving this problem. Various other solutions may lead to the same answers. This specific solution was provided by the Institute of Mathematical Sciences and Physics of the University of the Philippines Los Baños and explained by the University of the Philippines Civil Engineering Society.
PROBLEM:
For what value(s) of n is the series i + 2i2 + 3i3 + … + nin equal to 2000 + 2001i? Note that i is the imaginary unit.
SOLUTION: Seeing a series with increasing powers of i might have hinted you to search for a pattern in order to establish the given complex number, 2000 + 2001i. If so, then you’ve started off the same way we did. Through careful inspection, you will find that the given series can actually be grouped in such a way that you can define them as the sum of two series – one producing a real number and the other imaginary. Note that terms with even powers of i can be reduced to multiples of either 1 or -1 (i.e. real) and that terms with odd powers of i into multiples of either i or –i (i.e. imaginary). This was inferred from the property i2 = -1. With this established, let us begin by considering only the part of the series that would yield a real sum: 2i2 + 4i4 + 6i6 + … = -2 + 4 – 6 + 8 – 10 + ...
[1]
Look closely at equation [1]. Note that you can group the terms either like this: -2 + (4 – 6) + (8 – 10) + ... or like this (-2 + 4) + (-6 + 8) + (-10 + 12) +… To which you will find that when there is an odd number of terms, -2 + (4 – 6) + (8 – 10) + ... < 0, but we need 2000 > 0 This should alert you that the proper pairings should be: (-2 + 4) + (-6 + 8) + (-10 + 12) +… = 2 + 2 + … + 2 = 2000, where we have 1000 addends Hence, you’ll find that the sum of 2000 can be achieved with n = 4000 or n = 4001. Note that n = 4001 only affects the imaginary sum, while still yielding the real sum of 2000.
(Continued on next page)
UNIVERSITY OF THE PHILIPPINES CIVIL ENGINEERING SOCIETY Department of Civil Engineering College of Engineering and Agro-industrial Technology University of the Philippines Los Baños
(Continuation) To understand this logic, think about this for a second: If you have 1000 pairs of, say, red socks, you’d have 2000 red socks. But what if for every pair of red socks you buy, you get a pair of green socks? Then that would mean that you have 2000 red socks and 2000 green socks, which make a total of 4000 socks. Likewise, for every pairing of terms we made in the series above, there would always be a number before the first term of the pair and between the two terms of the pair (i.e. there is 1i before -2, and 3i3 between -2 and 4). With the real part done, let us now consider the part of the series that would yield an imaginary sum: 1i + 3i3 + 5i5 + … = 1i - 3i + 5i -7i + 9i + ...
[2]
Like in equation [1], you can group equation [2] in two different ways, but you’d have to choose the grouping that produces a sum of positive pairings for all n. This would be: 1i + (-3i + 5i) + (-7i + 9i) + … = 1i + (2i) + (2i) + … + (2i) = 2001i. Here you have a distinct term 1i and 1000 terms of 2i. Following the same logic from the real part of the series, we find that the sum 2001i can be achieved with n = 4001 or n = 4002. Note that n = 4002 only affects the real sum, while still yielding the imaginary sum of 2001i. Now, we have three values of n that could satisfy either of our conditions: n = 4000, n = 4001, and n = 4002, but of these three candidates, only n = 4001 satisfies both the conditions to produce the joint sum: 2000 + 2001i.
ANSWER:
n = 4001
UNIVERSITY OF THE PHILIPPINES CIVIL ENGINEERING SOCIETY Department of Civil Engineering College of Engineering and Agro-industrial Technology University of the Philippines Los Baños
