Get Answers to all your Questions

header-bg qa

If  X=\left \{ 4^{n}-3n-1\; :\; n\; \epsilon \; N \right \}\; and\;\; Y= \; \left \{9(n-1):n\; \epsilon \; N \right \},where\; N is the set of natural numbers,then

X\cup Y  is equal to :

Option: 1

X\;


Option: 2

Y\;


Option: 3

N\;


Option: 4

Y-X


Answers (1)

best_answer

Let us 1st simplify set X

X=(1+3)^n-3n-1

X=^nC_0+\;^nC_1\cdot3+\;^nC_2\cdot3^2\ldots \ldots+ \;^nC_n\cdot 3^n-3n-1X=1+\;3n+\;^nC_2\cdot3^2\ldots \ldots+ \; 3^n-3n-1

X=3^2\cdot\left (^nC_2+\;^nC_3\cdot3\ldots \ldots+ \; 3^{n-2} \right )

X=9\cdot\left (^nC_2+\;^nC_3\cdot3\ldots \ldots+ \; 3^{n-2} \right )

For n=1; X=0

For n=2; X=9

For n=3; X=54

:

:

And

Y=9\left ( n-1 \right )

For n=1; Y=0

For n=2; Y=9

For n=3; Y=18

:

For n=7; Y=54

:

Hence we can say

X\cup Y=Y

Posted by

Rishi

View full answer