 Hi children, my name is Mansi and I am going to help you solve the following question. The question says, prove the following by using the principle of mathematical induction for all n belonging to natural numbers. 1 plus 1 by 1 multiplied by 1 plus 1 by 2 multiplied by 1 plus 1 by 3 up till 1 plus 1 by n is equal to n plus 1. In this question we need to prove by using the principle of mathematical induction. Now before proving this we see the key idea behind the question. Now we know that the principle of mathematical induction is a specific technique which is used to prove certain statements that are formulated in terms of n where n is a positive integer. The principle can be explained with the help of two properties. If there is a given statement p at n such that first p at 1 is true and second if statement is true for n equal to k where k is some positive integer pk is true then statement p at k plus 1 is also true for n equal to k plus 1 then p at n is true for all natural numbers n. Now using these two properties we will show that statement is true for n equal to 1 then assume it is true for n equal to k then we prove that it is also true for n equal to k plus 1 hence proving that it is true for all n belonging to natural numbers. Now we start with the solution we have to prove that 1 plus 1 by 1 into 1 plus 1 by 2 into 1 plus 1 by 3 up till 1 plus 1 by n is equal to n plus 1. Now let p at n be 1 plus 1 by 1 into 1 plus 1 by 2 up till 1 plus 1 by n be equal to n plus 1. Now putting n equal to 1 p at 1 becomes 1 plus 1 by 1 that is 1 plus 1 and that is equal to 2 that is equal to 1 plus 1 and this is true. Now assuming that p at k is true p at k becomes 1 plus 1 by 1 into 1 plus 1 by 2 up till 1 plus 1 by k is equal to k plus 1 and let this be the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 becomes 1 plus 1 by 1 into 1 plus 1 by 2 into 1 plus 1 by 3 up till 1 plus 1 by k into 1 plus 1 by k plus 1 and this is same as k plus 1 into 1 plus 1 upon k plus 1 this we get using the first equation. Now adding the two expressions in the second bracket we get k plus 1 into k plus 1 plus 1 divided by k plus 1 that is equal to k plus 1 into k plus 2 upon k plus 1. For the we see that k plus 1 and k plus 1 they are common in numerator and denominator so they get cancelled and we left with k plus 2 that is equal to k plus 1 plus 1 and this is same as p at k plus 1. Thus p at k plus 1 is true wherever p at k is true. Hence from the principle of mathematical induction the statement p at n is true for all natural numbers n. So to solve this question we used the principle of mathematical induction we assumed that the statement is true for n equal to k and then we prove that n equal to k plus 1 is also true hence proved. I hope you understood the question and enjoyed the session. Goodbye.