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Prove by induction that 3n(n 1) is divisible by 6 for all positive integers.

User Yahya
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Answer with explanation:

We are asked to prove by the method of mathematical induction that:

3n(n+1) is divisible by 6 for all positive integers.

  • for n=1 we have:


3n(n+1)=3* 1(1+1)\\\\i.e.\\\\3n(n+1)=3* 2\\\\i.e.\\\\3n(n+1)=6

which is divisible by 6.

Hence, the result is true for n=1

  • Let the result is true for n=k

i.e. 3k(k+1) is divisible by 6.

  • Now we prove that the result is true for n=k+1

Let n=k+1

then


3n(n+1)=3(k+1)* (k+1+1)\\\\i.e.\\\\3n(n+1)=3(k+1)(k+2)\\\\i.e.\\\\3n(n+1)=(3k+3)(k+2)\\\\i.e.\\\\3n(n+1)=3k(k+2)+3(k+2)\\\\i.e.\\\\3n(n+1)=3k^2+6k+3k+6\\\\i.e.\\\\3n(n+1)=3k^2+3k+6k+6\\\\i.e.\\\\3n(n+1)=3k(k+1)+6(k+1)

Since, the first term:


3k(k+1) is divisible by 6.

( As the result is true for n=k)

and the second term
6(k+1) is also divisible by 6.

Hence, the sum:


3k(k+1)+6(k+1) is divisible by 6.

Hence, the result is true for n=k+1

Hence, we may say that the result is true for all n where n belongs to positive integers.

User Biberman
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