 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 into 2 plus 2 into 2 square plus 3 into 2 square up till n into 2 raise to the power n is equal to n minus 1 into 2 raise to the power n plus 1 plus 2. In this question we need to prove by using mathematical induction. Now before doing the solution to this question, 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 pk 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 and then we prove it is also true for n equals to k plus 1. Hence proving that it is true for all n belonging to natural numbers. Now we start with the proof of this question. Here we have to prove that 1 into 2 less 2 into 2 square plus 3 into 2 q up till n into 2 raise to the power n is equal to n minus 1 into 2 raise to the power n plus 1 plus 2. Let p at n be 1 into 2 plus 2 into 2 square plus 3 into 2 q and so on till n into 2 raise to the power n be equal to n minus 1 into 2 raise to the power n plus 1 plus 2. Now putting n equal to 1, p at 1 becomes 1 into 2 that is equal to 0 into 2 plus 2 that is 1 minus 1 into 2 raise to the power 1 plus 1 plus 2 that is equal to 2 and this is true. Now assuming that p at k is true p at k becomes 1 into 2 plus 2 into 2 square plus 3 into 2 q up till k into 2 k that is equal to k minus 1 into 2 raise to the power k plus 1 plus 2 and this becomes the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 is 1 into 2 plus 2 into 2 square plus 3 into 2 q up till k into 2 raise to the power k plus k plus 1 into 2 raise to the power k plus 1. This is equal to k minus 1 into 2 raise to the power k plus 1 plus 2 plus k plus 1 into 2 raise to the power k plus 1. This we get using first. This is same as k into 2 raise to the power k plus 1 minus 2 raise to the power k plus 1 plus 2 the whole plus k into 2 raise to the power k plus 1 plus 2 raise to the power k plus 1. This is same as 2 into k into 2 raise to the power k plus 1 plus 2. Now to represent the above expression in terms of k plus 1, it is further simplified k into 2 raise to power k plus 1 plus 1 plus 2 that is equal to k plus 1 minus 1 into 2 raise to power k plus 1 plus 1 plus 2. Now we see that 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 of mathematical induction the statement p at n is true for all natural numbers hence proved. I hope you understood the question and enjoyed the session. Goodbye.