UPSC NDA I 2018 Mathematics Solved Paper

Concept:The sum $(1+u)^n + (1-u)^n$ retains only terms with even powers of $u$.When $n$ is odd, the number of such terms is $\frac{n+1}{2}$.Explanation:We need the number of non‑zero terms in $(1+2\sqrt{3}x)^{11} + (1-2\sqrt{3}x)^{11}$ after simplification.Set $u = 2\sqrt{3}x$. Then the expression becomes $(1+u)^{11} + (1-u)^{11}$.Using the binomial theorem:$(1+u)^{11} = \sum\limits_{k=0}^{11} \binom{11}{k} u^k$$(1-u)^{11} = \sum\limits_{k=0}^{11} \binom{11}{k} (-u)^k = \sum\limits_{k=0}^{11} \binom{11}{k} (-1)^k u^k$When added, terms with odd $k$ cancel because $1+(-1)^k = 0$ for odd $k$.Even $k$ terms survive: $1+(-1)^k = 2$.Thus the sum equals $2\left[1 + \binom{11}{2}u^2 + \binom{11}{4}u^4 + \binom{11}{6}u^6 + \binom{11}{8}u^8 + \binom{11}{10}u^{10}\right]$.The powers of $u$ are $0,2,4,6,8,10$ — that is $6$ distinct terms.In general, for odd $n$, the number of non‑zero terms in $(1+u)^n+(1-u)^n$ is $\frac{n+1}{2}$.Here $n=11$, so $\frac{11+1}{2}=6$.Answer:6