 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 1 into 4 plus 1 divided by 4 into 7 plus 1 divided by 7 into 10 up till 1 divided by 3n minus 2 into 3n plus 1 is equal to n divided by 3n 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. 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 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 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. We have to prove that 1 upon 1 into 4 plus 1 upon 4 into 7 plus 1 upon 7 into 10 and so on till 1 upon 3n minus 2 into 3n plus 1 is equal to n upon 3n plus 1. Now let p at n be 1 divided by 1 into 4 plus 1 divided by 4 into 7 up till 1 divided by 3n minus 2 into 3n plus 1 let this be equal to n upon 3n plus 1. Now putting n equal to 1 p at 1 becomes 1 divided by 1 into 4 that is equal to 1 upon 4 that is equal to 1 upon 3 into 1 plus 1 and this is true. Now assuming that p at k is true p at k is 1 upon 1 into 4 plus 1 upon 4 into 7 plus 1 upon 7 into 10 up till 1 upon 3k minus 2 into 3k plus 1 and this is equal to k upon 3k plus 1 and this becomes the first equation. Now to prove that p at k plus 1 is also true. p at k plus 1 becomes 1 upon 1 into 4 plus 1 upon 4 into 7 up till 1 upon 3k minus 2 into 3k plus 1 plus 1 divided by 3 times k plus 1 minus 2 into 3 times k plus 1 plus 1. This is equal to k upon 3k plus 1 plus 1 upon 3 times k plus 1 minus 2 multiplied by 3 times k plus 1 plus 1 and this we get using first. Now 3 times k plus 1 minus 2 is equal to 3k plus 1 and 3 times k plus 1 plus 1 is equal to 3 times k plus 4. Thus the expression now becomes k upon 3k plus 1 plus 1 upon 3k plus 1 into 3k plus 4. Now adding the two expressions we get k into 3k plus 4 plus 1 divided by 3k plus 1 into 3k plus 4. This is equal to 3k square plus 4k plus 1 divided by 3k plus 1 into 3k plus 4. As 3k square plus 4k plus 1 is equal to 3k plus 1 into k plus 1 therefore this expression reduces to 3k plus 1 into k plus 1 divided by 3k plus 1 into 3k plus 4. Now we see that 3k plus 1 is common in numerator and denominator so they get cancelled and we get k plus 1 divided by 3k plus 4. Writing this result in terms of k plus 1 we get k plus 1 divided by 3 times k plus 1 plus 1. 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 hence proved. I hope you understood the question and enjoyed the session. Goodbye.