 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. 1Q plus 2Q plus 3Q up till nQ is equal to n into n plus 1, the whole divided by 2, the whole square. In this question we have to prove by using the principle of mathematical induction. Before proving this 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 p 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 can 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 cube plus 2 cube plus 3 cube up till n cube is equal to n into n plus 1 divided by 2 the whole square. Now let p at n be 1 cube plus 2 cube up till n cube is equal to n into n plus 1 by 2 the whole square. Now putting n equal to 1 p at 1 becomes equal to 1 cube that is same as n into n plus 1 by 2 the whole square when n equal to 1 it becomes 1 into 1 plus 1 divided by 2 the whole square. This is same as 2 divided by 2 and this is same as 1 and we see that this is true. Now assuming that p at k is true p at k becomes 1 cube plus 2 cube plus 3 cube up till k cube is equal to k into k plus 1 divided by 2 the whole square and this becomes the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 is equal to 1 cube plus 2 cube up till k cube plus k plus 1 the whole cube. This is equal to k into k plus 1 by 2 the whole square plus k plus 1 the whole cube and this we get using 1. This is same as k square into k plus 1 the whole square plus 4 into k plus 1 the whole cube the whole divided by 4. This is same as k square into k plus 1 the whole square plus 4 into k plus 1 the whole square into k plus 1 this whole divided by 4. Now we see that this can be written as k plus 1 the whole square into k square plus 4 k plus 4 the whole divided by 4. This becomes equal to k plus 1 the whole square into k plus 2 the whole square the whole divided by 4. This becomes equal to k plus 1 into k plus 2 divided by 2 the whole square. This is same as k plus 1 into k plus 1 plus 1 divided by 2 the whole square and we see that this is same as p at k plus 1. So 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.