If-n-is-a-multiple-of-4-and-i-1-find-the-sum-of-the-series-S-1-2i-3i-2-n-1-i-n-

Question Number 141585 by ZiYangLee last updated on 20/May/21

If n is a multiple of 4 and i = \sqrt{- 1},

find the sum of the series

S = 1 + 2 i + 3 i^{2} + \dots \dots + (n + 1) i^{n}

Answered by mathmax by abdo last updated on 21/May/21

S = 1 + 2 i + {3 i}^{2} + \dots . + (n + 1) i^{n} = \sum_{k = 0}^{n} (k + 1) i^{k}

we have \sum_{k = 0}^{n} x^{k} = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow \sum_{k = 1}^{n} {kx}^{k - 1} = \frac{d}{dx} (\frac{x^{n + 1} - 1}{x - 1})

= \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow \sum_{k = 0}^{n - 1} (k + 1) x^{k} \Rightarrow let change n by n + 1 \Rightarrow

\sum_{k = 0}^{n} (k + 1) x^{k} = \frac{(n + 1) x^{n + 2} - (n + 2) x^{n + 1} + 1}{{(x - 1)}^{2}}

if n = 4 p \Rightarrow \sum_{k = 0}^{n} (k + 1) i^{k} = \frac{(4 p + 1) i^{4 p + 2} - (4 p + 2) i^{4 p + 1} + 1}{{(i - 1)}^{2}}

= \frac{- (4 p + 1) - (4 p + 2) i + 1}{- 2 i} = \frac{4 p + 1 + (4 p + 2) i + 1}{2 i}

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

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