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$$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$$ where $4\mid n$. How can I simplify this exprerssion so as to obtain a general expression?

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    Is this a series or just a finite sum? – 2017-02-18
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    finite sum wherten isa multiple of 4 – 2017-02-18
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    powers of $i$ are such that for each $4$ steps you end up with the same result: $i^1 = i$,$i^2 = -1$,$i^3 = -i$,$i^4 = i^2i^2 = 1$, $i^5 = i = i^1$ – 2017-02-18
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    ihave figured that out but how will i simplify the whole sum as a single expression – 2017-02-18
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    Hint: replace $i$ with $x$, integrate, and solve. – 2017-02-18

4 Answers 4

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Hint...consider $$S-iS$$ and use the formula for the sum of a geometric series...

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    hmm niceway thanks ;) – 2017-02-18
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    You are welcome – 2017-02-18
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    That makes it too easy...LOL..........+1 – 2017-02-18
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    @user254665. Thanks, and its especially amusing since the geometric series part of it sums to zero... – 2017-02-18
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Consider functions $f(x) = 1 + 2x + 3x^2 +\ldots +(n+1)x^n$ and $F(x) = 1+ x + x^2 +\ldots + x^{n+1}$. We have $F'(x) = f(x)$. On the other hand, $F(x)(x-1) = x^{n+2} -1$ which implies $F(x) = \frac{x^{n+2}-1}{x-1}$ for $x\neq 1$. From this we have $$f(x) = F'(x) = \frac{(n+2)x^{n+1}(x-1)-(x^{n+2}-1)}{(x-1)^2}$$ All you have to do is evaluate at $i$.

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    Except when i=1...........................+1 – 2017-02-18
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Hint.

You have $i^2=-1$, $i^3=-i$ and $i^4=1$.

So because $n$ is a multiple of $4$:

$$S=\sum_{k=0}^n (k+1)i^k=\sum_{k=0}^{n/4} 4(k+1)+i\sum_{k=0}^{n/4} (4(k+1)+1)-\sum_{k=0}^{n/4} (4(k+1)+2)-i\sum_{k=0}^{n/4} (4(k+1)+3).$$

You now just have to simplify all the four sums.

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Let $n=4m$.

$$\begin{align} S&=\boxed{\begin{array} &\;\;\;\;1&+2i&+3i^2&+4i^3\\ +5i^4&+6i^5&+7i^6&+8i^7\\ +\vdots\\ +(4m-3)i^{4m-4}&+(4m-2)i^{4m-3}&-(4m-1)i^{4m-2}&+4mi^{4m-1}\\ +(4m+1)i^{4m}\\ \end{array}}\\\\ &=\boxed{\begin{array} &\;\;\;1&+2i&-3&-4i\\ +5&+6i&-7&-8i\\ +\vdots\\ +(4m-3)&+(4m-2)i&-(4m-1)&-4mi\\ +(4m+1)\\ \end{array}}\\\\ &=\;\;\;\;\;\;[1-3+5-7+\cdots+(4m-3)-(4m-1)]\\ &\;\;\;\;+i\;[2-4+6-8+\cdots +(4m-2)-4m]\\ &\;\;\;\;\;\;\;\;\;+(4m+1)\\ &=[\overbrace{-2-2-\cdots-2}^m]+i[\overbrace{-2-2\cdots-2}^m]+(4m+1)\\ &=-2m+i(-2m)+(4m+1)\\ &=2m+1-i2m\\ &=\left(\frac n2+1\right)-i\frac n2 \end{align}$$