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Question

If A={x:x=4 power n minus 3n minus 1 and n belongs to mode of N } and B = {y:y = 9(n -1) and n belongs to mode of N } Prove that A subset of B

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Solution

This can be proved by mathematial induction.
Let x=4^n-3n-1
y=9(m-1).

For n=0, x=0.
For m=1, y=0.
Which means the element 0 exists in A as well as B.

Let us assume x=some y, that is
4^n-3n-1 = 9(m-1) for some n and m.
So if we prove for the value of x at (n+1), we can show there is some y with y=9(k-1), where k belongs to N, then we hav proved it.

4^n-3n-1 = 9(m-1)
4^n-1 = 9(m-1)+3n-------------I

For n+1,
4^(n+1) - 3(n+1) -1
= 4x4^n - 3n- 4
=4(4^n-1) - 3n
= 4[9(m-1)+3n)] - 3n (From I)
= 36(m-1) -9n
= 9[4(m-1)-n)
=9[(4m-n-3)-1)
=9(k-1) where k=4m-n-3 which again belongs to N.

Hence Proved.

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