Question:
Prove that $\mathrm{i}^{53}+\mathrm{i}^{72}+\mathrm{i}^{93}+\mathrm{i}^{102}=2 \mathrm{i}$
Solution:
L.H.S $=\mathrm{i}^{53}+\mathrm{i}^{72}+\mathrm{i}^{93}+\mathrm{i}^{102}$
$=i^{4 \times 13+1}+i^{4 \times 18}+i^{4 \times 23+1}+i^{4 \times 25+2}$
Since $i^{4 n}=1$
$\Rightarrow i^{4 n+1}=i$ (where $n$ is any positive integer)
$\Rightarrow i^{4 n+2}=-1$
$\Rightarrow i^{4 n+3}=-1$
$=i+1+i+i^{2}$
$=i+1+i-1$
$=2 \mathrm{i}$
L.H.S = R.H.S
Hence proved.
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