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Question

Prove that for any positive interger n,n3n is divisible by 6.

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Solution

Let f(n)=n3n.
Now f(1)=0, divisible by 6.
Again f(2)=6, divisible by 6.
So the given statement is true for n=1,2.
Let us take the statement is true for some integer k.
f(k)=k3k is divisible by 6.
Let f(k)=6k1 [k1 being an integer].......(1).
Now
f(k+1)
=(k+1)3(k+1)
=k3+3k2+2k
=k3k+3k2=3k
=(k3k)+3k(k+1).........(2).
k(k+1) is the product of two consecutive integers, which is always divisible by 2.
Then we can write k(k+1)=2k2 [Where k2 is some integer.].
Using this and (1) form (2) we get,
f(k+1)=6k1+6k2=6(k1+k2)
k1,k2 are integers then k1+k2 is also an integer.
So f(k+1) is also divisible by 6.
So the statement is true for n=k+1 if we assume it to be true for n=k.
By the principle of mathematical induction the statement is true nN.

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