 Hi children, my name is Mansi and I am going to help you solve the following question. The question here says, proof the following by using the principle of mathematical induction for all n belonging to natural numbers, x raised to the power of 2n minus y raised to the power of 2n is divisible by x plus y. In this question we need to prove by using the principle of mathematical induction. Now before starting the proof 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 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 x raised to the power of 2n minus y raised to the power of 2n is divisible by x plus y. Let p at n be x raised to the power of 2n minus y raised to the power of 2n is divisible by x plus y. Putting n equal to 1 p at 1 becomes x raised to the power of 2 into 1 minus y raised to the power of 2 into 1 is equal to x square minus y square that is same as x plus y into x minus y. Now x plus y into x minus y is divisible by x plus y thus p at 1 is true. Now assuming that p at k is true p at k becomes x raised to the power of 2k minus y raised to the power of 2k is divisible by x plus y. We write that x raised to the power of 2k minus y raised to the power of 2k is equal to x plus y multiplied by d 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 x raised to the power of 2 into k plus 1 minus y raised to the power of 2 into k plus 1 Now as 2 into k plus 1 is equal to 2k plus 2 so x raised to the power of 2 into k plus 1 minus y raised to the power of 2 into k plus 1 is same as x raised to the power of 2k plus 2 minus y raised to the power of 2k plus 2. Now adding and subtracting x raised to the power of 2 and y raised to the power of 2k this expression becomes equal to x raised to the power of 2k plus 2 minus x squared into y raised to the power of 2k plus x squared into y raised to the power of 2k minus y raised to the power of 2k plus 2. Now as x raised to the power of 2k plus 2 is equal to x raised to the power of 2k x squared and y raised to the power of 2k plus 2 is equal to y raised to the power of 2k into y squared. So this becomes equal to x raised to the power of 2k into x squared minus x squared into y raised to the power of 2k plus x squared into y raised to the power of 2k minus y raised to the power of 2k into y squared. This is equal to x squared into x raised to the power of 2k minus y raised to the power of 2k plus y raised to the power of 2k into x squared minus y squared. This is same as x squared into x plus y multiplied by d plus y raised to the power of 2k into x plus y into x minus y and this we get using 1. Now taking x plus y as common expression becomes equal to x plus y multiplied by x squared d plus y raised to the power of 2k into x minus y. The above expression is divisible by x plus y. Thus x raised to the power of 2 into k plus 1 minus y raised to the power of 2 into k plus 1 is divisible by x plus y. 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. I hope you understood the question and enjoyed the session. Goodbye.