\( \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]\)
Note that
i1 = i
i2 = -1
i3 = i * i2 = i * -1 = - i
i4 = i2 * i2 = -1 * - 1 = 1
So....the sum of these = 0
And this pattern continues for each partial sum of
i1+4n, i2+4n, i3 + 4n and i4 + 4n where n is an integer ≥ 0
So......we will have
0 + 0 + 0 + .......+ 0 + 0 + i97 + i98 + i99 =
i1+ 4(24) + i 2 + 4(24) + i 3 + 4(24) =
[ i ] + [ -1 ] + [ - i ] =
-1
Can someone help me I do not understand the question
Simplify \(i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\)
Geometric sequence ratio = i
\(\begin{array}{|rcll|} \hline s_{99} &=& \frac{ i - i^{100} } {1-i} \qquad & | \qquad i^{100} = i^{2\cdot 50} = (i^2)^{50} = (-1)^{50} = 1 \\\\ s_{99} &=& \frac{ i - 1 } {1-i} \\\\ s_{99} &=& - \left( \frac{ 1 - i } {1-i} \right) \\\\ s_{99} &=& - 1 \\ \hline \end{array} \)