UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The powers of $i$ repeat every 4 terms, and the sum of any four consecutive powers is zero.Explanation:We need $\sum\limits_{n=1}^{8n+7} i^n$.Here the upper limit is $8n+7$, where $n$ is a positive integer.The sum of a complete cycle of four powers is $i^1+i^2+i^3+i^4 = i -1 -i +1 = 0$.Total number of terms is $8n+7$.Divide $8n+7$ by $4$: $8n+7 = 4(2n+1) + 3$.This means there are $(2n+1)$ full groups of four (each sum $0$) and $3$ leftover terms.The leftover terms are $i^{8n+5}$, $i^{8n+6}$, and $i^{8n+7}$.Since $i^{4k}=1$, we have:$i^{8n+5} = i^{4(2n+1)+1} = i^1 = i$$i^{8n+6} = i^{4(2n+1)+2} = i^2 = -1$$i^{8n+7} = i^{4(2n+1)+3} = i^3 = -i$Sum of leftovers: $i + (-1) + (-i) = -1$.Answer:$-1$