Respuesta :
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:
[tex]3n(n+1)=3\times 1(1+1)\\\\i.e.\\\\3n(n+1)=3\times 2\\\\i.e.\\\\3n(n+1)=6[/tex]
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
[tex]3n(n+1)=3(k+1)\times (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)[/tex]
Since, the first term:
[tex]3k(k+1)[/tex] is divisible by 6.
( As the result is true for n=k)
and the second term [tex]6(k+1)[/tex] is also divisible by 6.
Hence, the sum:
[tex]3k(k+1)+6(k+1)[/tex] 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.
