 Hi children, my name is Mansi and I am 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 a plus ar plus ar squared up till ar raise to power n minus 1 equal to a into r raise to power n minus 1 the whole divided by r minus 1. In this question we need to prove by using the principle of mathematical induction. Now before doing the 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 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 a plus ar plus ar square up till ar n minus 1 is equal to a into rn minus 1 divided by r minus 1. Now let p at n be a plus ar plus ar square up till ar raise to power n minus 1 equal to a into r raise to power n minus 1 divided by r minus 1. Now putting n equal to 1 p at 1 is a that is a into r raise to power 1 minus 1 divided by r minus 1 that is equal to a into r minus 1 upon r minus 1 the same as a and this is true. Now assuming that p at k is true p at k becomes a plus ar plus ar square up till ar k minus 1 that is a into r raise to power k minus 1 divided by r minus 1 and this is the first equation. Now to prove that p at k plus 1 is also true p at k plus 1 becomes a plus ar square up till a into r raise to power k minus 1 plus a into r raise to power k. This is equal to a into r raise to power k minus 1 divided by r minus 1 plus a into r raise to power k and this we get using first. Now adding the two expressions we get a into r raise to power k minus 1 plus a into r raise to power k into r minus 1 this whole divided by r minus 1. It is the same as a into r raise to power k minus a plus a into r raise to power k plus 1 minus a into r raise to power k divided by r minus 1. Now putting a into r raise to power k minus a into r raise to power k is equal to 0 and a into r raise to power k plus 1 minus a is equal to a into r raise to power k plus 1 minus 1. It becomes a into r raise to power k plus 1 minus 1 the whole divided by r minus 1 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. So to solve this question we use the principle of mathematical induction. We assume the statement to be true for n equal to k then proved that n equal to k plus 1 is also true. Hence proved. Hope you enjoyed the session. Goodbye.