 Hi children, my name is Mansi and I'm going to help you with 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 2 into 5 plus 1 divided by 5 into 8 plus 1 divided by 8 into 11 up till 1 divided by 3n minus 1 into 3n plus 2 is equal to n divided by 6n plus 4. In this question we need to prove by using the principle of mathematical induction. Now before starting the solution 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 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. Now we have to prove that 1 divided by 2 into 5 plus 1 divided by 5 into 8 plus 1 divided by 8 into 11 up till 1 divided by 3n minus 1 into 3n plus 2 is equal to n divided by 6n plus 4. Let p at n be 1 divided by 2 into 5 plus 1 divided by 5 into 8 up till 1 divided by 3n minus 1 into 3n plus 2 is equal to n divided by 6n plus 4. Now putting n equal to 1 p at 1 becomes 1 divided by 2 into 5 is equal to 1 divided by 10 that is same as 1 divided by 6 into 1 plus 4 and we see that this is true. Now assuming that p at k is true p at k becomes 1 divided by 2 into 5 plus 1 divided by 5 into 8 up till 1 divided by 3k minus 1 into 3k plus 2 is equal to k divided by 6k plus 4 this becomes the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 becomes 1 divided by 2 into 5 plus 1 divided by 5 into 8 plus 1 divided by 8 into 11 and so on till 1 divided by 3k minus 1 into 3k plus 2 plus 1 divided by 3 times k plus 1 minus 1 multiplied by 3 times k plus 1 plus 2. Now this is equal to k divided by 6k plus 4 plus 1 divided by 3 times k plus 1 minus 1 multiplied by 3 times k plus 1 plus 2 this we get using 1 this is equal to k divided by twice of 3k plus 2 plus 1 divided by 3k plus 2 into 3k plus 5. Now adding the two expressions this becomes k times 3k plus 5 plus 2 this divided by twice of 3k plus 2 into 3k plus 5 this is equal to 3k square plus 5k plus 2 divided by twice of 3k plus 2 into 3k plus 5. Now this can also be written as 3k square plus 3k plus 2k plus 2 divided by twice of 3k plus 2 into 3k plus 5 this is equal to 3k plus 2 into k plus 1 divided by twice of 3k plus 2 into 3k plus 5 now we see here that 3k plus 2 is common enumerator and denominator so they get cancelled and what we get is k plus 1 divided by 6k plus 10. Now representing the expression in terms of k plus 1 we get k plus 1 divided by 6 times k plus 1 plus 4 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 at n is true for all natural numbers n so to solve this question we use the principle of mathematical induction we assume the statement to be true for n equal to k and then prove that n equal to k plus 1 is also true hence proved I hope you understood the question and enjoyed the session goodbye