 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question here says, prove the following by using the principle of mathematical induction for all n belonging to natural numbers, n into n plus 1 into n plus 5 is a multiple of 3. Let us start with the solution to this question. We have to prove here that n into n plus 1 into n plus 5 is a multiple of 3. Let this be p at n. Now putting n equal to 1, p at 1 becomes 1 into 1 plus 1 into 1 plus 5 equals to 1 into 2 into 6 that is equal to 12. We see that 12 is a multiple of 3 and this is true because 3 into 4 equals to 12 and this makes 12 a multiple of 3. Now assuming that p at k is true, p at k is k into k plus 1 into k plus 5 is a multiple of 3. We write k into k plus 1 into k plus 5 is equal to 3d where d belongs to n and let this be the first equation. Now to prove that v at k plus 1 is also true putting n equal to k plus 1 we find k plus 1 into k plus 1 plus 1 into k plus 1 plus 5 is same as k plus 1 into k plus 2 into k plus 6 that is equal to k plus 1 into k square plus 8k plus 12. Now multiplying the two brackets we get k cube plus 8k square plus 12k plus k square plus 8k plus 12. This is equal to k cube plus 9k square plus 20k plus 12. Now writing the equation in such a form to get 3d, we can write it as k cube plus 6k square plus 3k square plus 5k plus 15k plus 12. This can be written as k into k square plus 6k plus 5 plus 3 into k square plus 5k plus 4. We see this is equal to k into k plus 1 into k plus 5 plus 3 into k square plus 5k plus 12. This is equal to 3d plus 3 into k square plus 5k plus 12. Let this we get using 1. This is equal to 3 into d plus k square plus 5k plus 12 and we see that this expression is a multiple of 3. Thus e at k plus 1 is true. Hence from the principle of mathematical induction the statement p at n is true for all natural numbers hence proved. So I hope you understood the question and enjoyed the session. Have a good day.