Question:

If \[ \sum_{k=0}^{440} i^k=x+iy, \] then \[ x^{100}+x^{99}y+x^{242}y^2+x^{97}y^3= \]

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The powers of \(i\) repeat every \(4\) terms: \[ 1,\; i,\; -1,\; -i \] Use this cyclic property to simplify large powers and summations involving \(i\).
Updated On: Jun 22, 2026
  • \(0\)
  • \(-4\)
  • \(4\)
  • \(1\)
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The Correct Option is D

Solution and Explanation

Step 1: Use the cyclic property of powers of \(i\).
The powers of \(i\) repeat in cycles of \(4\): \[ i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i \] and then the pattern repeats.
Also, \[ 1+i-1-i=0 \] Hence, every block of \(4\) consecutive powers sums to zero.

Step 2: Simplify the summation.
Given, \[ \sum_{k=0}^{440} i^k \] Since \[ 440=4\times110, \] the terms from \(i^0\) to \(i^{439}\) form \(110\) complete cycles whose sum is zero.
Thus only the last term remains: \[ i^{440} \] Now, \[ i^{440}=(i^4)^{110}=1^{110}=1 \] Therefore, \[ x+iy=1 \] Comparing real and imaginary parts, \[ x=1,\qquad y=0 \]

Step 3: Substitute the values of \(x\) and \(y\).
We need to evaluate \[ x^{100}+x^{99}y+x^{242}y^2+x^{97}y^3 \] Substituting \[ x=1,\qquad y=0, \] we get \[ 1^{100}+1^{99}(0)+1^{242}(0)^2+1^{97}(0)^3 \] \[ =1+0+0+0 \] \[ =1 \]

Step 4: Final conclusion.
Hence, \[ \boxed{1} \] which corresponds to option (4).
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