Question

show that i^m+i^m+1+i^m+2+i^m+3=0 for all m€N

Answer 1

The answer is given below :

Now, L.H.S.

$= {i}^{m} + {i}^{m + 1} + {i}^{m + 2} + {i}^{m + 3} \\ \\ = {i}^{m} (1 + i + {i}^{2} + {i}^{3} ) \\ \\ = {i}^{m} (1 + i - 1 - i) \\ \\ = {i}^{m} \times 0 \\ \\ = 0$
for all m belongs to |N.

= R.H.S. [Proved]

$here \: \: i = \sqrt{( - 1)} \\ \\ so \: \: {i}^{2} = - 1 \\ \\ and \\ \\ {i}^{3} \\ \\ = i \times {i}^{2} \\ \\ = i \times ( - 1) \\ \\ = - i$

Thank you for your question.