7.1, page 342, exercises 3, 11, and 24. #24. By replacing n by n 1 in (7.1.14), write a formula for Sn1. Subtract the formula for Sn 1from the formula for Sn and use the result to derive the recurrence relation Sn = 2Sn 1 Sn 2 + Sn 3. section 7.2, page 356, exercises 5 solve the given recurrence relation for the initial conditions given. 5. Exercise 4;a0 = 0 Section 7.1, page 342, exercises 3, 11, and 24. Please only answer question 3 and 11 on chart below and the onter 2 questions that needs to be answered are below #24. By replacing n by n 1 in (7.1.14), write a formula for Sn1. Subtract the formula for Sn 1from the formula for Sn and use the result to derive the recurrence relation Sn = 2Sn 1 Sn 2 + Sn 3. section 7.2, page 356, exercises 5 Tell whether or not each recurrence relation in Exercises 1-10 is a linear homogeneous recurrence relation with constant coefficients. Give the order of each linear homogeneous recurrence relation with constant coefficients. 5.an = 7an 2 6an 3 11. At the end of n-1 years, the amount is A n1 . After one more year, we will have the amount A n1 plus the interest. Thus A n= A n1+ ( 0.1 ) A n1 =(1.1) A n1 , The value of n 1. A 0 is the initial value that is the beginning amount. A 0=2000 So, Now, A 1=( 1.1 ) A 0=2200 A 2=( 1.1 ) A 1=2420 A 3=( 1.1 ) A 2=2662 These are the desired values. 24. Substituting n-1 for n in the recurrence relation of equation 7.1.14 that is the equation S n=S n1 + Sn 3 + Sn 4 +...+ S1 +3 We get S n1=Sn2 + Sn4 + S n5+ ...+ S2 + S1 +3 The expression for S n and S n1 overlap in the sum S n4 + S n5 + ...+S 2+ S 1+ 3 . Taking the difference, we find that S nS n1 =( S n1 +S n3 )S n2 => S n=S n1 + ( S n1+ S n3 ) Sn2 => S n=2 Sn1 +S n3S n2 5. an =7 an26 an3 Yes, this recurrence relation is a linear homogenous recurrence relation with constant coefficients. The, order of this linear homogenous recurrence relation with constant coefficients is 3. Answer 1) A key is the variable which is used for the purpose of encryption or decryption. A key is used at the sender address to encrypt the message and the key is used at the receiver end to decrypt the encrypted message. 2) Public and private keys are keys which are used to encrypt and decrypt the data at sender and receiver end. The public key is available to everyone and the private key is available to the authorized receiver. The encryption in the RSA system is done with the help of private key while the key generation is completed with the help of public keys. 3) Xx 4) Xx 5) Security is used everywhere, from mobile device to computer systems, one business network system to another business network system. Everyone wants to secure his/her data from theft and unauthorized person, so security is very important for everyone. No one wants that their message gets leaked to unauthorized people and delivered to the receiver in a different form than the original message. So, in order to keep the data and information safe and secure, the security is required. The security is also required to maintain the confidentiality, integrity and availability of the data