 Hi children, my name is Mansi and I'm 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 divided by 1 plus 2 plus 1 divided by 1 plus 2 plus 3 up till 1 divided by 1 plus 2 plus 3 up till n equals to 2n by n plus 1. In this question we need to prove using the principle of mathematical induction. Now before starting with the solution we see the key idea behind the question. 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 of n such that first p of 1 is true and second if statement is true for n equal to k where k is some positive integer p of k is true then statement p at k plus 1 is also true for n equal to k plus 1 then p of n is true for all natural numbers. 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 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 to this question we have to prove that 1 plus 1 divided by 1 plus 2 plus 1 divided by 1 plus 2 plus 3 up till 1 divided by 1 plus 2 plus 3 up till n is equal to 2n divided by n plus 1. Let p of n be 1 plus 1 divided by 1 plus 2 plus 1 divided by 1 plus 2 plus 3 up till 1 divided by 1 plus 2 plus 3 up till n and this is equal to 2n divided by n plus 1 putting n is equal to 1 p at 1 becomes 1 which is equal to 2 multiplied by 1 divided by 1 plus 1 that is equal to 2 divided by 2 which is equal to 1 and this is true. Now assuming that p at k is true p at k becomes 1 plus 1 divided by 1 plus 2 plus 1 divided by 1 plus 2 plus 3 up till 1 divided by 1 plus 2 plus 3 up till k is equal to 2k divided by k plus 1. Now to prove that p at k plus 1 is also true p at k plus 1 is 1 plus 1 divided by 1 plus 2 plus 1 divided by 1 plus 2 plus 3 up till 1 divided by 1 plus 2 plus 3 up till k plus 1 divided by 1 plus 2 plus 3 up till k plus 1 is equal to 2k divided by k plus 1 plus 1 divided by 1 plus 2 plus 3 up till k plus 1. Now we know that sum of natural numbers n is equal to 1 plus 2 plus 3 up till n is equal to n multiplied by n plus 1 divided by 2 we put 1 plus 2 plus 3 up till k plus 1 is equal to k plus 1 multiplied by k plus 1 plus 1 the whole divided by 2 we get equal to 2k divided by k plus 1 plus 2 divided by k plus 1 multiplied by k plus 1 plus 1 that is equal to 2k multiplied by k plus 2 plus 2 the whole divided by k plus 1 multiplied by k plus 2 that is equal to twice of k square plus 2k plus 1 divided by k plus 1 multiplied by k plus 2 that is equal to twice of k plus 1 multiplied by k plus 1 divided by k plus 1 multiplied by k plus 2 this we get because k square plus 2k plus 1 is equal to k plus 1 multiplied by k plus 1 this is equal to twice of k plus 1 divided by k plus 2 this is equal to twice of k plus 1 divided by k plus 1 plus 1 this is equal to p of k plus 1 thus p of k plus 1 is true wherever p at k is true hence from the principle of mathematical induction the statement p of n is true for all natural numbers n hence proved I hope you understood the question and enjoyed the session goodbye