UPSC NDA I 2018 Mathematics Solved Paper

Concept:The sum of powers of $i$ repeats in cycles of 4:$i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, and then the pattern repeats.A key property: $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for any integer $n$.Explanation:We need to compute $\sum_{n=2}^{11} (i^n + i^{n+1})$.Step 1: Write the sum explicitly:$S = (i^2+i^3) + (i^3+i^4) + (i^4+i^5) + \cdots + (i^{10}+i^{11}) + (i^{11}+i^{12})$.Step 2: Notice that each term $i^k$ for $k=3$ to $11$ appears twice, while $i^2$ and $i^{12}$ appear only once.So $S = i^2 + 2(i^3 + i^4 + \cdots + i^{11}) + i^{12}$.Step 3: The middle sum $i^3 + i^4 + \cdots + i^{11}$ contains 9 terms. Group them into sets of 4:$(i^3+i^4+i^5+i^6) + (i^7+i^8+i^9+i^{10}) + i^{11}$.Each group of 4 consecutive powers sums to 0, so only $i^{11}$ remains.Thus $i^3 + \cdots + i^{11} = i^{11}$.Step 4: Substitute back:$S = i^2 + 2 \cdot i^{11} + i^{12}$.Now compute each power:$i^2 = -1$,$i^{11} = i^{4\cdot2 + 3} = (i^4)^2 \cdot i^3 = 1^2 \cdot (-i) = -i$,$i^{12} = i^{4\cdot3} = (i^4)^3 = 1^3 = 1$.Step 5: Therefore,$S = -1 + 2(-i) + 1 = -1 - 2i + 1 = -2i$.Answer:The sum equals $-2i$, which corresponds to option C.