 Hi children, my name is Mansi and I am 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 upon 1 into 2 into 3 plus 1 upon 2 into 3 into 4 plus 1 upon 3 into 4 into 5 up till 1 upon n into n plus 1 into n plus 2 is equal to n into n plus 3 divided by 4 into n plus 1 into n plus 2. In this question we need to prove by using the principle of mathematical induction. Now before doing the question 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 and 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 to this question. In this question we have to prove that 1 upon 1 into 2 into 3 plus 1 upon 2 into 3 into 4 plus 1 upon 3 into 4 into 5 and so on till 1 upon n into n plus 1 into n plus 2 is equal to n into n plus 3 divided by n plus 1 into n plus 2 and this whole multiplied by 4. Now let p at n be 1 upon 1 into 2 into 3 plus 1 upon 2 into 3 into 4 plus 1 upon 3 into 4 into 5 and so on till 1 upon n into n plus 1 into n plus 2 is equal to n into n plus 3 divided by 4 into n plus 1 into n plus 2. Now putting n equal to 1 p at 1 becomes 1 upon 1 into 2 into 3 is equal to 1 upon 1 multiplied by 1 plus 1 multiplied by 1 plus 2 and this is equal to 1 upon 6 and this is true. Now assuming that p at k is true p at k becomes 1 upon 1 into 2 into 3 plus 1 upon 2 into 3 into 4 plus 1 upon 3 into 4 into 5 and so on till 1 upon k into k plus 1 into k plus 2 is equal to k into k plus 3 divided by 4 into k plus 1 into k plus 2. Let this be the equation 1. Now we have to prove that p at k plus 1 is also true. P at k plus 1 is 1 upon 1 into 2 into 3 plus 1 upon 2 into 3 into 4 plus 1 upon 3 into 4 into 5 and so on till 1 upon k into k plus 1 into k plus 2 plus 1 upon k plus 1 into k plus 2 into k plus 3. Now we see that this is equal to k into k plus 3 divided by 4 into k plus 1 into k plus 2 plus 1 upon k plus 1 into k plus 2 into k plus 3 and this we get using first. Now adding the two expressions we get k into k plus 3 into k plus 3 plus 4 divided by 4 into k plus 1 into k plus 2 into k plus 3. This is same as k cube plus 6 k square plus 9 k plus 4 divided by 4 into k plus 1 into k plus 2 into k plus 3. Now as k cube plus 6 k square plus 9 k plus 4 can be written as k cube plus 2 k square plus 4 k square plus k plus 8 k plus 4. This becomes equal to k cube plus 2 k square plus 4 k square plus k plus 8 k plus 4 this entire term divided by 4 into k plus 1 into k plus 2 into k plus 3. Now taking k square plus 2 k plus 1 as common we get k into k square plus 2 k plus 1 plus 4 into k square plus 2 k plus 1. This entire term divided by 4 into k plus 1 into k plus 2 into k plus 3. This is equal to k plus 4 into k square plus 2 k plus 1 divided by 4 into k plus 1 into k plus 2 into k plus 3. Now as k plus 1 into k plus 1 is equal to k square plus 2 k plus 1 so we get this to be equal to k plus 4 into k plus 1 into k plus 1 divided by 4 into k plus 1 into k plus 2 into k plus 3. Now cancelling k plus 1 we get k plus 1 into k plus 4 divided by 4 into k plus 2 into k plus 3. Representing the expression in terms of k plus 1 we get k plus 1 into k plus 1 plus 3 divided by 4 into k plus 1 plus 1 into k plus 1 plus 2 which 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 hence proved. I hope you understood the question and enjoyed the session goodbye.