Question

Show that i^m + i^m+1 + i^m+2 + i^m+3 = 0 for all m belongs to N.​

Answer 1

Step-by-step explanation:

$4 {i}^{m} + 1 + 2 + 3 = 0$

$4 {i}^{m} = -6$

${i}^{m} = \frac{ - 6}{4}$

${i}^{m} = \frac{ - 3}{2}$

substitute the value in the question

L.H.S

$= 4( \frac{ - 3}{2} ) + 1 + 2 + 3$

$= 2( - 3) + 1 + 2 + 3$

$= - 6 + 1 + 2 + 3$

$= - 6 + 6$

$= 0$

L.H.S=R.H.S

Hence Proved