Question

$find \: the \: value \: of \\ \\ \: \: i + {i}^{2} + {i}^{3} + {i}^{4} + .... + {i}^{202}$
Chapter : complex number ​

Answer 1

Step-by-step explanation:

Given:

$i + {i}^{2} + {i}^{3} + {i}^{4} + .... + {i}^{202}$

To find:Find the value

Solution:

We know that

${i}^{2} = - 1 \\ \\ {i}^{3} = - i \\ \\ {i}^{4} = 1$

Now divide the given terms in pair of four

$(i + {i}^{2} + {i}^{3} + {i}^{4} ) + ( {i}^{5} + {i}^{6} + {i}^{7} + {i}^{8}) \\ \\ + ... + ( {i}^{197} + {i}^{198} + {i}^{199} + {i}^{200}) + {i}^{201} + {i}^{202}$

Place the values

$(i - 1 - i + 1) + ( {i}^{4 + 1} + {i}^{4 + 2} + {i}^{4 + 3} + {i}^{4 \times 2} ) + ... + ( {i}^{4 \times 49 + 1} + {i}^{4 \times 49 + 2} + {i}^{4 \times 49 + 3} + {i}^{4 \times 50} ) + {i}^{4 \times 50 + 1} + {i}^{4 \times 50 + 2}$

All the four pair will result to zero as the first pair

$(i - 1 - i + 1) + (i - 1 - i + 1) +... + (i - 1 - i + 1) + i - 1$

$0 + 0 + ... + 0 + i - 1$

$i - 1$

Final answer:

$\boxed{\bold{i + {i}^{2} + {i}^{3} + {i}^{4} + .... + {i}^{202}=i-1}}$

Hope it helps you.

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