Help again?

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i.................... 1 + i - 1 - i = - 1. Sorry for the typo!!.

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

Simplify $$i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.$$

$\begin{array}{|rcll|} \hline && \displaystyle i^1+i^2+i^3+i^4+i^5+\cdots+ i^{97} + i^{98}+i^{99} \\\\ &=& (i^1+i^2+i^3+i^4)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(1+i^2)\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \quad & | \quad i^2 = -1 \\\\ &=& (i^1+i^2)(\underbrace{1+i^2}_{=0})\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \\\\ &=& 0-i^{100} \\\\ &=& -(i^2)^{50} \\\\ &=& -(-1)^{50} \\\\ &=& -1 \\ \hline \end{array}$