Prove that

Question:

Prove that $\frac{1}{i}-\frac{1}{i^{2}}+\frac{1}{i^{3}}-\frac{1}{i^{4}}=0$

 

Solution:

Given: $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}$

To prove : $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}=0$.

$\Rightarrow$ L.H.S. $=\mathrm{i}^{-1}-\mathrm{i}^{-2}+\mathrm{i}^{-3}-\mathrm{i}^{-4}$

$\Rightarrow \mathrm{i}^{-4 \times 1+3}-\mathrm{i}^{-4 \times 1+2}+\mathrm{i}^{-4 \times 1+3}-\mathrm{i}^{-4 \times 1}$

Since $i^{4 n}=1$

$\Rightarrow i^{4 n+1}=i$

$\Rightarrow \mathrm{i}^{4 \mathrm{n}+2}=-1$

$\Rightarrow i^{4 n+3}=-1$

So,

$\Rightarrow \mathrm{i}^{1}-\mathrm{i}^{2}+\mathrm{i}^{3}-1$

$\Rightarrow \mathrm{i}+1-\mathrm{i}-1$

$\Rightarrow 0$

$\Rightarrow$ L.H.S $=$ R.H.S

Hence Proved

 

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