Question
Download Solution PDFWhat is \(\rm \sum\limits_{n=1}^{8n+7} i^n\) equal to, where i = √-1?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFFormula used:
Powers of iota.
i2 = -1, i3 = -i, i4 = 1,
i5 = ( i . i4 ) = i,
i6 = i2 . i4 = -1,
i7 = -i...... and so on.
⇒ i1 + i2 + i3 + i4 = i -1 - i + 1 = 0
Calculation:
\(\rm \sum\limits_{n=1}^{8n +7} i^n\)
⇒ i1 + i2 + i3 + i4 + i5 + i6 + i7 + ..... + i8n+7
⇒ i1 + i2 + i3 + i4 + i5 + i6 + i7 + ..... + i8n + i8n+1 + i8n+2 + .... + i8n+6 + i8n+7
⇒ (i -1 - i + 1) + (i -1 - i + 1) + .... + (i8n+1 + i8n+2 + i8n+3 + i8n+4 ) + (i8n+5 + i8n+6 + i8n+7)
⇒ 0 + i8n+5 + i8n+6 + i8n+7 = i1 + i2 + i3 = i -1 - i = -1
∴ The value of \(\rm \sum\limits_{n=1}^{8n +7} i^n\) is - 1.
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