 Hello and welcome to the session. My name is Mansi and I'm 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 into 3 plus 3 into 5 plus 5 into 7 up to 2n minus 1 into 2n plus 1 equals to n into 4n square plus 6n minus 1 the whole divided by 3. Let us start with the solution to this question. Here we have to prove that 1 into 3 plus 3 into 5 plus 5 into 7 up till 2n minus 1 into 2n plus 1 equals to n into 4n square plus 6n minus 1 divided by 3. Let p at n be 1 into 3 plus 3 into 5 plus 5 into 7 up till 2n minus 1 into 2n plus 1 is equal to n into 4n square plus 6n minus 1 the whole divided by 3. Putting n equal to 1 p at 1 becomes 1 into 3 that is same as 9 divided by 3 that is equal to 4 plus 6 minus 1 divided by 3 which is equal to 1 into 4 into 1 square plus 6 into 1 minus 1 divided by 3 which is equal to 3 and we see that this is true. Now assuming that p at k is true we have p at k is 1 into 3 plus 3 into 5 and so on till 2k minus 1 into 2k plus 1 is equal to k into 4k square plus 6k minus 1 divided by 3. Let this be the first equation. Now to prove that p at k plus 1 is also true let p at k plus 1 be 1 into 3 plus 3 into 5 plus 5 into 7 till 2k minus 1 into 2k plus 1 plus twice of k plus 1 minus 1 into 2 into k plus 1 plus 1 which is same as k into 4k square plus 6k minus 1 divided by 3 plus twice of k plus 1 minus 1 into 2 k plus 1 plus 1 this we get using the first equation which is same as 4k cube plus 6k square minus k divided by 3 plus 2k plus 1 into 2k plus 3 we see that this is equal to 4k cube plus 6k square minus k divided by 3 plus 4k square plus 8k plus 3. Now this can be further written as 4k cube plus 6k square minus k plus 3 into 4k square plus 8k plus 3 divided by 3 this is equal to 4k cube plus 18k square plus 23k plus 9 divided by 3. Now to represent the above expression in the terms of k plus 1 it is further simplified as 4k cube plus 14k square plus 4k square plus 9k plus 14k plus 9 divided by 3 this is equal to k into 4k square plus 14k plus 9 plus 1 into 4k square plus 14k plus 9 the whole divided by 3. Now taking k plus 1 as common multiple we get k plus 1 into 4k square plus 14k plus 9 divided by 3 now to represent the second bracket in terms of k plus 1 it is again simplified as k plus 1 into 4k square plus 8k plus 6k plus 4 plus 5 divided by 3 we see that this is equal to k plus 1 into 4 into k plus 1 the whole square plus 6k plus 5 divided by 3 now representing the whole expression in terms of k plus 1 we get k plus 1 into 4 into k plus 1 the whole square plus 6 into k plus 1 minus 1 divided by 3 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 n hence proved so i hope you understood the question and enjoyed the session have a good day