 Hi children, 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 3 raised to power 2n plus 2 minus 8n minus 9 is divisible by 8. In this question we need to prove by using the principle of mathematical induction. Now before proving this 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 a solution to this question. Here we have to prove that 3 raised to power 2n plus 2 minus 8n minus 9 is divisible by 8. Let p at n be 3 raised to power 2n plus 2 minus 8n minus 9 is divisible by 8. Putting n equal to 1 p at 1 becomes 3 raised to power 2 into 1 plus 2 minus 8 into 1 minus 9 is equal to 3 raised to power 4 minus 8 minus 9 that is same as 64. Now 64 is divisible by 8 thus p at 1 is true. Now assuming that p at k is true p at k becomes 3 raised to power 2k plus 2 minus 8k minus 9 is divisible by 8. Now we write 3 raised to power 2k plus 2 minus 8k minus 9 as 8d where d belongs to the natural numbers 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 have to find 3 raised to power 2k plus 1 plus 2 minus 8 into k plus 1 minus 9. This is same as 3 raised to power 2k plus 2 plus 2 minus 8k minus 8 minus 9. Now adding and subtracting 3 square into 8k and 3 square into 9 expression becomes equal to 3 raised to power 2k plus 2 plus 2 minus 3 square into 8k plus 3 square into 8k minus 8k plus 3 square into 9 minus 3 square into 9 minus 70. Now grouping the expressions according to common multiples we get 3 raised to power 2k plus 2 plus 2 minus 3 square into 8k minus 3 square into 9 plus 3 square into 8k minus 8k plus 3 square into 9 minus 70. Now taking the common multiples out of the brackets we get 3 square into 3 raised to power 2k plus 2 minus 8k minus 9 plus 8k into 3 square minus 1 plus 3 square into 9 minus 70. This is same as 3 square into 8d plus 8k into 8 plus 64. This we get using the first equation this is equal to 8 into 9d plus 8k plus 8. Now the above expression is divisible by 8 thus 3 raised to power 2 into k plus 1 plus 2 minus 8 into k plus 1 minus 9 is divisible by 8 thus p at k plus 1 is true. Hence from the principle of mathematical induction the statement at n is true for all natural numbers hence proved. I hope you understood the question and enjoyed the session. Goodbye.