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).