 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. 1 plus 3 by 1 the whole multiplied by 1 plus 5 by 4 the whole multiplied by 1 plus 7 by 9 up till 1 plus 2n plus 1 divided by n square is equal to n plus 1 the whole square. 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 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 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 plus 3 by 1 into 1 plus 5 by 4 up till 1 plus 2 n plus 1 divided by n square is equal to n plus 1 the whole square. Let p at n be 1 plus 3 by 1 multiplied by 1 plus 5 by 4 multiplied by 1 plus 7 by 9 up till 1 plus 2 n plus 1 by n square is equal to n plus 1 the whole square. Now putting n equal to 1 p at 1 becomes 1 plus 3 by 1 is equal to 1 plus 2 into 1 plus 1 by 1 square that is equal to 1 plus 3 by 1 and this is true. Now assuming that p at k is true p at k becomes 1 plus 3 by 1 into 1 plus 5 by 4 into 1 plus 7 by 9 up till 1 plus 2 k plus 1 upon k square is equal to k plus 1 the whole square and this becomes the first equation. Now we have to prove that p at k plus 1 is also true. p at k plus 1 becomes 1 plus 3 by 1 into 1 plus 5 by 4 up till 1 plus 2 k plus 1 by k square multiplied by 1 plus twice of k plus 1 plus 1 divided by n square divided by k plus 1 the whole square. Now we see that this is same as k plus 1 the whole square plus 1 plus twice of k plus 1 plus 1 this divided by k plus 1 the whole square plus 1 this divided by k plus 1 the whole square and this we get using first. Now adding the two expressions in the second bracket this becomes equal to k plus 1 the whole square k plus 1 the whole square multiplied by k plus 1 the whole square plus 2 k plus 3 divided by k plus 1 the whole square. This is same as k plus 1 the whole square multiplied by k square plus 1 plus 2 k plus 2 k plus 3 divided by k plus 1 the whole square. This is same as k plus 1 the whole square multiplied by k square plus 4 k plus 4 divided by k plus 1 the whole square. Also k square plus 4 k plus 4 the same as k plus 2 the whole square so this becomes equal to k plus 1 the whole square multiplied by k plus 2 the whole square divided by k plus 1 the whole square. Now k plus 1 the whole square is common in both numerator and denominator it is cancelled. Therefore we are left with k plus 2 the whole square which is same as k plus 1 plus 1 the whole square. Now 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 that the principle of mathematical induction statement p 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 is equal to k plus 1 is also true hence proved. I hope you understood the question and enjoyed the session. Goodbye.