 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 squared plus 3 squared plus 5 squared up till 2n minus 1 the whole square equals to n into 2n minus 1 into 2n plus 1 the whole divided by 3. 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 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. We have to prove that 1 square plus 3 square plus 5 square up till 2n minus 1 the whole square is equal to n into 2n minus 1 into 2n plus 1 this whole divided by 3. Let p at n be 1 square plus 3 square plus 5 square up till 2n minus 1 the whole square is equal to n into 2n minus 1 into 2n plus 1 this whole divided by 3. Now, putting n equal to 1 p at 1 becomes 1 square that is 1 that is equal to 1 into 2 into 3 upon 3. This is equal to 1 into 1 plus 1 into 1 plus 2 divided by 3 and this is also equal to 1 so this is true. Now assuming that p at k is true p at k be 1 square plus 3 square plus 5 square up till 2k minus 1 the whole square is equal to k into 2k minus 1 into 2k plus 1 the whole divided by 3 and this becomes the first equation. Now to prove that p at k plus 1 is also true. p at k plus 1 be 1 square plus 3 square plus 5 square up till 2k minus 1 the whole square plus twice of k plus 1 minus 1 the whole square. This is equal to k into 2k minus 1 into 2k plus 1 divided by 3 plus twice of k plus 1 minus 1 the whole square and this we get using first. Now twice of k plus 1 minus 1 is equal to 2k plus 1 thus we get k into 2k minus 1 into 2k plus 1 divided by 3 plus 2k plus 1 the whole square. Now adding the two expressions this becomes k into 2k minus 1 into 2k plus 1 plus 3 into 2k plus 1 the whole square this whole divided by 3. Now taking 2k plus 1 as a common multiple this can be written as 2k plus 1 multiplied by k into 2k minus 1 plus 3 into 2k plus 1 this whole divided by 3. That is equal to 2k plus 1 into 2k square minus k plus 6k plus 3 this whole divided by 3. Now as 2k square plus 6k minus k plus 3 is equal to 2k plus 3 into k plus 1 this becomes equal to 2k plus 1 into 2k plus 3 into k plus 1 this whole divided by 3. Now writing this result in terms of k plus 1 we get k plus 1 into twice of k plus 1 minus 1 into twice of k plus 1 plus 1 this divided by 3. That 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 enjoyed the session and understood the question goodbye.