Find the value of $1 + i^{2} + i^{4} + i^{6} + i^{8} + . . . . . + i^{20}$ .

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Solution

$1 + i^{2} + i^{4} + i^{6} + . . . \dots . + I^{20}$
This is a GP with $a = 1$ & $r = i^{2} = - 1$
$i^{20} = (1) (i^{2})^{n - 1}$
$\Rightarrow i^{2 \times 10} = i^{2 (n - 1)}$
So $n - 1 = 10$
$\Rightarrow n = 11$
$S_{11} = \frac{(1) (1 - r^{n})}{(1 - r)} = \frac{1 - (- 1)^{11}}{1 - (- 1)} = \frac{1 - (- 1)}{1 - (- 1)} = 1$ .

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Find the value of 1+i2+i4+i6+i8+.....+i20.

1+i2+i4+i6+...….+I20This is a GP with a=1 & r=i2=−1i20=(1)(i2)n−1⇒i2×10=i2(n−1)So n−1=10⇒n=11S11=(1)(1−rn)(1−r)=1−(−1)111−(−1)=1−(−1)1−(−1)=1.

Find the value of $1 + i^{2} + i^{4} + i^{6} + i^{8} + . . . . . + i^{20}$ .