 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, 10 raised to power 2n minus 1 plus 1 is divisible by 11. In this question we need to prove by using the principle of mathematical induction. Before starting the solution, 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 the 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 to this question. Here we have to prove that 10 raised to power 2n minus 1 plus 1 is divisible by 11. Now let p at n be 10 raised to power 2n minus 1 plus 1 is divisible by 11. Now putting n equal to 1, p at 1 becomes 10 raised to power 2 into 1 minus 1 plus 1 is equal to 10 plus 1 that is equal to 11. Now we see that 11 is divisible by 11, thus p at 1 is true. Now assuming that p at k is true, p at k is 10 raised to power 2k minus 1 plus 1 is divisible by 11. We write 10 raised to power 2k minus 1 plus 1 is equal to 11d where d belongs to n and this becomes the first equation. Now to prove that p at k plus 1 is also true, putting n equal to k plus 1 we find 10 raised to power twice of k plus 1 minus 1 plus 1 as twice of k plus 1 minus 1 is equal to 2k plus 1 we get 10 raised to power 2k plus 1 plus 1 that is equal to 10 raised to power 2k minus 1 plus 2 plus 1 as 1 is same as 100 minus 99 and 10 square is equal to 100 thus we get 10 raised to power 2k minus 1 into 10 square plus 100 minus 99. This is same as 10 square multiplied by 10 raised to power 2k minus 1 plus 1 minus 99 this is equal to 11d into 10 square minus 11 into 9 and this we get using first. It is the same as 11 into d multiplied by 10 square minus 9 this expression is divisible by 11 thus 10 raised to power twice of k plus 1 minus 1 plus 1 is divisible by 11 thus p at k plus 1 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 goodbye.