 Hello and welcome to the session. In this session we will discuss the concept of sequences and recurrence relationships. Now consider the following pattern of bells. Here you can see first layer has 1 bell, second layer has 3 bells, third layer has 1, 2, 3, 4 and 5 bells, third layer has 7 bells. Now let a n represents number of bells in nth layer. So number of bells in first layer are given by even which is equal to 1. Number of bells in second layer are given by a2 which is equal to 3. Number of bells in third layer are given by a3 which is equal to and number of bells in fourth layer are given by a4 which is equal to 7. Now if the pattern is continued we get a sequence of numbers 1, 3, 5, 7. The next number will be 9, then 11, 13, 15, 17, 21. Here the string of bells shows that pattern continues forever. This is a function whose domain is a set of all positive integers. Now in this sequence of bells we are getting all all numbers starting from 1. Now any general sequence is written as a1, a2, a3, a4 and so on and a n, a n plus 1, a n plus 2 and so on where a n is the first term of the sequence, a n is the nth term of the sequence, even term plus one of term given by a n plus 1. Now let us see types of sequence. Now there are three types of sequences, arithmetic sequence, geometric sequence and third is Fibonacci sequence. Now let us see what is recurrence relationship. Now recurrence relationship is a formula which helps us in finding all the terms of the sequence, some terms or facts about the sequence. Now in this section we will discuss recurrence relationships for arithmetic sequence and Fibonacci sequence. Now let us see what is a Fibonacci sequence. Now the most famous example of a recurrence relationship is that which generates Fibonacci sequence 1, 1, 2, 3, 5, 13, 21, 34, 55 and so on. Now this pattern starts with 2 after this each term is obtained by adding the two terms preceding it. Now here first term of the sequence is a1 which is equal to 1 and here the second term that is a2 is equal to 1. Now here the third term that is a3 is equal to a2 which is equal to 1 plus 1 which is equal to 2. So here a3 so here this term is obtained by adding a4 will be equal to a3 plus a2 which is equal to 2 plus 1 that is equal to 3. Thus a5 is equal to 3 similarly a5 that is fifth term will be equal to fourth term that is a term that is a3. Now a4 is equal to a3 is equal to 2 so a5 will be equal to 3 plus 2 which is equal to 5. Now a6 will which is equal to 5 plus 3 that is equal to is equal to plus a1 that is two preceding terms relationship for Fibonacci sequence. Then we get plus 2 is equal to a1 plus 1 plus a1 which implies a1. So we have a3 is equal to now a2 here is 1 and a1 is again 1 of the sequence is equal to 2 3 4 5 and so on. And here we will take all positive integers obtain all the terms of the sequence. Now consider the above sequence of words. You can identify the pattern from given terms. Now here first term is a1 which is equal to 1. Second term that is a2 is equal to 3. Third term that is a3 is equal to 5. Fourth term that is a4 is equal to 7. That is a5 is equal to 9 and so on. We can see that a2 that is 3 minus 1 is equal to 2. Then a3 minus a2 is equal to 5 minus 3 which is again 2. Then is equal to 7 which is also 2. Now here we see difference from the previous term by 2. It means difference is common is called common difference which is denoted by sequence is a sequence in which is from the previous term by the same fixed number. Now we will derive the alternative sequence when the first term and common difference are known using reference relation is equal to d, a3 minus a2 is equal to d, d is equal to d and if we continue this pattern plus 1 minus am is equal to d which implies am plus 1 is equal to am and this is the reference relation for alternative sequence. The value of d that is the common difference that is the first term of the sequence then we can find all other terms in the sequence using this relation. Now here n is a positive integer so we will put n is equal to 1, 2, 3, 4, 5 and so on obtain other terms of the sequence. Now suppose we have a1 is equal to 3 and d is equal to 4 denote this equation by equation 1 and now we will put n is equal to 1 in equation 1 and here we have a1 plus 1 is equal to a1 plus d which implies a2 is equal to a1 plus d, a1 is 3 and d is 4. So this implies a2 is equal to 3 plus 4, a2 is equal to 7 and now let us put n is equal to 2 in equation 1 and we have a2 plus 1 is equal to a2 plus d which implies a3 is equal to now a2 is equal to 7 plus d that is 4 and this implies a3 is equal to 11, 3 plus d which is equal to 11 plus 4 which is equal to 15, 2, a3, a4 and so on using reckons now a1 is 3, a2 is 7, then a3 is 11, then a4 is 15 and so on. So in this session we have discussed the concept of sequences and reckons relationships and this concludes our session. Hope you all have enjoyed the session.