Question:
Prove that $\left(1+\mathrm{i}^{2}+\mathrm{i}^{4}+\mathrm{i}^{6}+\mathrm{i}^{8}+\ldots .+\mathrm{i}^{20}\right)=1$
Solution:
L.H.S $=\left(1+i^{2}+i^{4}+i^{6}+i^{8}+\ldots .+i^{20}\right)$
$\sum_{n=0}^{n=20} i^{n}$
$=1+-1+1+-1+\ldots \ldots \ldots . .+1$
As there are 11 times 1 and 6 times it is with positive sign as ${ }^{i}{ }^{0}=1$ as this is the extra term and there are 5 times 1 with negative sign.
So, these 5 cancel out the positive one leaving one positive value i.e. 1
$\sum_{n=0}^{20} i^{n}=1$
L.H.S = R.H.S
Hence proved.
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