 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 divided by 3 into 5 plus 1 divided by 5 into 7 plus 1 divided by 7 into 9 up till 1 divided by 2n plus 1 into 2n plus 3 is equal to n divided by 3 into 2n plus 3. In this question we need to prove by 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 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 p at k 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. 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. Here we have to prove that 1 divided by 3 into 5 plus 1 divided by 5 into 7 plus 1 divided by 7 into 9 up till 1 divided by 2n plus 1 into 2n plus 3 is equal to n divided by 3 into 2n plus 3. Now let p at n be 1 divided by 3 into 5 plus 1 divided by 5 into 7 up till 1 divided by 2n plus 1 into 2n plus 3 is equal to n divided by 3 into 2n plus 3. Now putting n equal to 1 p at 1 becomes 1 divided by 3 into 5 that is same as 1 divided by 2 plus 3 that can be written as 2 into 1 plus 3 and we see that this is true. Now assuming that p at k is true p at k becomes 1 divided by 3 into 5 plus 1 divided by 5 into 7 up till 1 divided by 2k plus 1 into 2k plus 3 is equal to k divided by 3 into 2k plus 3. This becomes the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 is 1 divided by 3 into 5 plus 1 divided by 5 into 7 and so on till 1 divided by 2k plus 1 into 2k plus 3 plus 1 divided by twice of k plus 1 plus 1 into twice of k plus 1 plus 3. This is equal to k divided by 3 into 2k plus 3 plus 1 divided by twice of k plus 1 plus 1 multiplied by twice of k plus 1 plus 3 and this we get using first. Now we know that twice of k plus 1 plus 1 is equal to 2k plus 3 and twice of k plus 1 plus 3 is 2k plus 5. Therefore this expression becomes equal to k divided by 3 into 2k plus 3 plus 1 divided by 2k plus 3 into 2k plus 5. Now adding the two expressions we get k into 2k plus 5 plus 3 divided by 3 times 2k plus 3 into 2k plus 5. This is equal to 2k square plus 5k plus 3 divided by 3 times 2k plus 3 into 2k plus 5. Now as 2k square plus 5k plus 3 is equal to 2k plus 3 into k plus 1 so this expression now becomes 2k plus 3 into k plus 1 divided by 3 multiplied by 2k plus 3 into 2k plus 5. Now taking 2k plus 3 as common multiple 2k plus 3 gets cancelled and we are left with k plus 1 divided by 3 multiplied by 2k plus 5. Now writing this result in terms of k plus 1 we get k plus 1 divided by 3 multiplied by twice of k plus 1 plus 3 and we see that this is 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 hence proved. I hope you understood the question and enjoyed the session. Goodbye.