+85 The answer is -1.
let's first start with small numbers, i^1 + i^2 + i^3 + i^4. Assuming in your case i is the imaginary number, I'll start by calculating that. i^1 is just i. i^2 is -1, i^3 is -1(i) which is -i, and i^4 is -i(i) which is -i^2 which is -1(-1) which becomes 1. Continiuing to calculate these, you'll find that there's a pattern.
i^1 = i i^5 = i
i^2 = -1 i^6 = -1
i^3 = -i i^7 = -i
i^4 = 1 i^8 = 1
The powers of i repeat in groups of 4.
Now that we've found out the pattern, we'll calculate yours. 99/4 leaves a remainder of 3. So then we have i^97 = i, i^98 = -1, and i^99 = -1. The i and -i cancel out, so you're left with the answer -1.
+26388 Find i^1+i^2+i^3+...+ i^97 + i^98+i^99.
\(\begin{array}{|rcll|} \hline s &=& i^1+i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99} \\ i*s &=& i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99}+i^{100} \\ \hline s-i*s &=& i^1 - i^{100} \\\\ s(1-i) &=& i - i^{100} \\\\ s &=& \dfrac{i - i^{100}}{1-i} \\\\ s &=& \dfrac{i - i^{2*50}}{1-i} \\\\ s &=& \dfrac{ i - \left(i^2\right)^{50} }{1-i} \quad | \quad i^2 = -1 \\\\ s &=& \dfrac{ i - \left(-1\right)^{50} }{1-i} \\\\ s &=& \dfrac{ i - 1 }{1-i} \\\\ s &=&- \dfrac{ (1 - i) }{(1-i)} \\\\ \mathbf{s} &=& \mathbf{-1} \\\\ \mathbf{i^1+i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99} } &=& \mathbf{-1} \\ \hline \end{array}\)
