# 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

Akshay
185 Points
8 years ago
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 *3n-2) = 9*z, where z is an integer,
z will only have specific integer values. Hence zUN = N, or XUY=Y
ATB
Akshay
185 Points
8 years ago
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 *3n-2) = 9*z, where z is an integer,
z will only have specific integer values. Hence zUN = N, or XUY=Y
ATB
Prateeti Sengupta
28 Points
8 years ago
Could you please clarify your steps with reasons??? I ddin’t get what you meant...
Akshay
185 Points
8 years ago
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.
manmath
49 Points
7 years ago
) = 9*z, where z is an integer,z will only have specific integer values. Hence zUN = N, or XUY=YATBn-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