Let ω=e^2π i/5. List-I (P) Σk=1^4|1-ω^k|^2 (Q) prodk=1^4(1-ω^k (R) Σk=0^4(1+ω^k^7 (S) prodk=0^4(2-ω^k List-II (1) 5 (2) 10 (3) 110 (4) 31 (5) 0

(P) $|1-\omega^{k}|^{2}=2-2\cos\frac{2\pi k}{5}$, and $\sum_{k=1}^{4}\cos\frac{2\pi k}{5}=-1$, so $P=8+2=10$, which is (2). (Q) The roots of $\frac{x^{5}-1}{x-1}=1+x+x^{2}+x^{3}+x^{4}$ are $\omega^{k}$, $k=1,\dots,4$; evaluating at $x=1$ gives $\prod(1-\omega^{k})=5$, which is (1). (R) Expanding, $\sum_k(1+\omega^{k})^{7}=\sum_j\binom7j\sum_k\omega^{jk}$ and the inner sum is $5$ when $5\mid j$ and $0$ otherwise, so $R=5\left[\binom70+\binom75\right]=5(1+21)=110$, which is (3). (S) Since $\prod_{k=0}^{4}(x-\omega^{k})=x^{5}-1$, putting $x=2$ gives $31$, which is (4). Hence P→2, Q→1, R→3, S→4. Numerical evaluation with complex arithmetic gives $10$, $5$, $110$, $31$, all matching. Option (5) is unused. Notice that all four parts use the identity $x^{5}-1=\prod_{k}(x-\omega^{k})$ or the orthogonality of roots of unity. A frequent error in (R) is to include more binomial terms than the multiples of five, or to forget the factor five from the sum over $k$.