X=4n−3n−1:n∈N
4n−3n−1=(3+1)n−3n−1
=nC0(3)o1n+nC1(3)1(1)n−1+.....+nCn(3)n−3n−1
=1+n(3)+nC232+.....+3n−3n−1
=nC232+nC333+.....+3n
=32(nC2+nC331+nC432+.....+3n−2)
=9k, where k is a natural number.
Y=9(n−1):n∈N
∴X⊆Y
∴X∪Y=Y
JEE Main 2014 — Mathematics Algebra
If X=4n−3n−1:n∈N and Y=9(n−1):n∈N, where N is the set of natural numbers, then X∪Y is equal to
Held on 6 Apr 2014 · Verified 6 Jul 2026.
X
Y
N
Y−X
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\}$, is :
Suppose $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. and $\mathrm{a}^{2}, 2 \mathrm{~b}^{2}, \mathrm{c}^{2}$ are in G.P. If $\mathrm{a}<\mathrm{b}<\mathrm{c}$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$, then $9\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)$ is equal to $\_\_\_\_$ .
Let $S=\left\{z \in \mathbb{C}: 4 z^{2}+\bar{z}=0\right\}$. Then $\sum_{z \in S}|z|^{2}$ is equal to:
Let $S=\{z: 3 \leqslant|2 z-3(1+i)| \leqslant 7\}$ be a set of complex numbers. Then $\min _{z \in S}\left|\left(z+\frac{1}{2}(5+3 i)\right)\right|$ is equal to :
Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{bmatrix}$ and $B = [b_{ij}]$, $1 \leq i, j \leq 3$. If $B = A^{99} - I$, then the value of $\dfrac{b_{31} - b_{21}}{b_{32}}$ is :
Work through every JEE Main Algebra PYQ, year by year.