If X= {4n – 3n -- 1:n belongs to N} and Y= {9(n-1): n belongs to N}, where N is the set of Natural numbers, prove that X U Y = Y

4ⁿ=(3+1)ⁿ = nC0 + nC1*3¹ + nC2 * 3² + nC3 * 3³ +........+ nCn * 3ⁿ,
Take nC0 and nC1*3 on left hand side,
x=nC2 *9 + nC3*27...... + nCn *3ⁿ,
hence x is always divisible by 9,
So you can write x= 9*(nC2 + nC3*3......+ nCn *3^n-2) = 9*z, where z is an integer,
z will only have specific integer values. Hence zUN = N, or XUY=Y
ATB

4ⁿ=(3+1)ⁿ = nC0 + nC1*3¹ + nC2 * 3² + nC3 * 3³ +........+ nCn * 3ⁿ,
Take nC0 and nC1*3 on left hand side,
x=nC2 *9 + nC3*27...... + nCn *3ⁿ,
hence x is always divisible by 9,
So you can write x= 9*(nC2 + nC3*3......+ nCn *3^n-2) = 9*z, where z is an integer,
z will only have specific integer values. Hence zUN = N, or XUY=Y
ATB

Could you please clarify your steps with reasons??? I ddin’t get what you meant...

OK.
If set A is a subset of set B, then AUB=B. In this question, set Y={9(n-1)}. Y={0,9,18,27.....}
In other words, Y is a set of elements that are divisible by 9.
Set X={4ⁿ-3n-1}. To prove XUY=Y, we have to first prove that X is a set of numbers that are divisible by 9.
For that, you have to expand 4ⁿ, by using binomial theorm. Just write 4=(3+1). 
 4n=(3+1)n = nC0 + nC1*31 + nC2 * 32 + nC3 * 33 +........+ nCn * 3n,
When you put this in 4ⁿ-3n-1, 3n+1 will get cancelled out. So, X={nC2 *3² + nC3*3³...... + nCn *3n
}.
Now X will always be divisible by 9, as each term in expanded version of X has minimum power of 3².
Y is a set of all natural numbers that are divisible by 9. So, any set which contains integers that are divisible by 9, will be a subset of Y, as Y is a universal set of atural numbers divisible by 9. Hence, XUY=Y.
ATB.

) = 9*z, where z is an integer,z will only have specific integer values. Hence zUN = N, or XUY=YATB^n-2,hence x is always divisible by 9,So you can write x= 9*(nC2 + nC3*3......+ nCn *3ⁿ,Take nC0 and nC1*3 on left hand side,x=nC2 *9 + nC3*27...... + nCn *3ⁿ +........+ nCn * 3³ + nC3 * 3² + nC2 * 3¹ = nC0 + nC1*3ⁿ=(3+1)⁴ⁿ