PLEASE SHOW FULL WORKING HAS ME STUMPED FOR EXAM REVISION Th

PLEASE SHOW FULL WORKING HAS ME STUMPED FOR EXAM REVISION

The sequence of terms 1, 3, 9, 27, , 3 where n is a non-negative integer (n > 0), is called a geometric progression with common ratio 3. Use the principle of mathematical induction to show that the sum of the first n + 1 terms of this geometric progression is equal to3n + 1-1/2. In other words, prove the following formula by induction 1 + 3 + 3 2 + + 3n = 3n + 1-1/2. Use the above result to calculate

Solution

check for n = 1 1+3 = 9-1/2 hence true now assume true for n = k then 1+3+9.......+3^k = (3^k+1 - 1 )/2 add 3^k+1 on both sides we have rhs as (3^k+1 - 1 )/2 + 3^k+1 = (3^k+1 - 1 + 2.3^k+1)/2 =(3^k+1 - 1 + (3-1).3^k+1)/2 =(3^k+1 - 1 + 3^k+2 - 3^k+1 )/2 = (3^k+2 - 1 )/2 hence we showed that it is true for n = k+1 hence by mathematical induction it is true for any n b)substitute n = 15 in that formula to get the result hence answer is (3^15 - 1 )/2