Simplify \(i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\)
\(\displaystyle \sum \limits_{k=1}^n~\alpha^k =\dfrac{\alpha \left(\alpha ^n-1\right)}{\alpha -1}\)
letting \(\alpha=i,~n=99,~\text{and assuming }i=\sqrt{-1}\)
\(\displaystyle \sum \limits_{k=1}^{99} ~i^k = \dfrac{i(i^{100}-1)}{i-1} = 0\)
Note the repeating pattern of length 4
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
So....the sum of this series from i^1 to i*96 = 0
So...the sum of the last three terms = i^97 + i^98 + i^99 = i + -1 + -i = -1
So...this series sums to 0 + (-1) = -1
Simplify
\(\large{i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.} \)
Geometric sequence:
\(\begin{array}{|rcll|} \hline s &=& i^1+&i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} \quad & | \quad \cdot i \\ i\cdot s &=& &i^2+i^3+i^4+\cdots+ i^{98} + i^{99}+i^{100} \\\\ \hline s-i\cdot s &=& i^1 - i^{100} \\ s(1-i) &=& i^1 - i^{100} \\\\ \mathbf{s} &\mathbf{=} & \mathbf{ \dfrac{i - i^{100}}{1-i} } \quad & | \quad i^{100}= \left(i^2 \right)^{50}=(-1)^{50}=1 \\\\ s & = & \dfrac{i-1}{1-i} \\ s & = & -\left( \dfrac{1-i}{1-i} \right) \\ \mathbf{s} &\mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)
\(\boxed{\large{i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} = -1}}\)