Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342. 8.2 Solving First-Order Linear Recurrence EquationsA first-order linear recurrence equation relates consecutive entries in a sequenceby an equation of the formSn+1 = aS, +cfor Vn in the domain of S.(8.2.1)But let's assume that the domain of S is N. Let's also assume that a + 0; otherwise,S, = c for Vn 0, and the solutions to (8.2.1) are not very interesting.I/ What are they?We saw in Chap. 3, that when a = 1, any sequence satisfying (8.2.1) is anarithmetic sequence, and when S is defined on N and So is some initial value I,S, = 1+ncfor Vn e N.// Theorem 3.6.4Also, when e = 0, any sequence satisfying (8.2.1) is a geometric sequence, andwhen S is defined on N and So is some initial value I,S, = d"Ifor Vn e N.I/ Theorem 3.6.7Furthermore, Theorem 3.6.8 gives a formula for the sum of the first (n + 1) termsof a geometric series when a + 1:a°1+a'I+a²I+.. +a'I = 1(1+a+a² +... +a") = 1ªа - 1 Therefore, the general solution of the recurrence equationSp+1 = aS, +cfor Vn eN(8.2.1)is given in two parts:if a = 1,S, = 1+ncfor Vn e N;if a + 1,S„ = a"A +•for Vn e N.When a = 1, any particular solution is obtained by determining a specific,numerical value for I. In fact, a particular solution is determined by a specific,numerical value J for any (particular) entry, S,. Solving the equationJ = I+jc for I,1 = J – je.I/ since S; = I+jc// where So =1we get// One particular “particular solution" has I = 0.When a + 1, any particular solution is obtained by determining a specific,numerical value for A; if the starting value I is given, then A = 1-,In fact,1- aa particular solution is determined by a specific, numerical value J for any(particular) entry, S,. Solving the equationJ = Ad +for A,awe getA =// But what if a = 0?// One particular “particular solution" has A = 0.

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Solve the first-order linear recurrence relation: Sn+1 = 5 Sn + 1, with S0=1. You may use the general solution given on P.342. 8.2 Solving First-Order Linear Recurrence EquationsA first-order linear recurrence equation relates consecutive entries in a sequenceby an equation of the formSn+1 = aS, +cfor Vn in the domain of S.(8.2.1)But let's assume that the domain of S is N. Let's also assume that a + 0; otherwise,S, = c for Vn > 0, and the solutions to (8.2.1) are not very interesting.I/ What are they?We saw in Chap. 3, that when a = 1, any sequence satisfying (8.2.1) is anarithmetic sequence, and when S is defined on N and So is some initial value I,S, = 1+ncfor Vn e N.// Theorem 3.6.4Also, when e = 0, any sequence satisfying (8.2.1) is a geometric sequence, andwhen S is defined on N and So is some initial value I,S, = d&#34;Ifor Vn e N.I/ Theorem 3.6.7Furthermore, Theorem 3.6.8 gives a formula for the sum of the first (n + 1) termsof a geometric series when a + 1:a°1+a'I+a²I+.. +a'I = 1(1+a+a² +... +a&#34;) = 1ªа - 1 Therefore, the general solution of the recurrence equationSp+1 = aS, +cfor Vn eN(8.2.1)is given in two parts:if a = 1,S, = 1+ncfor Vn e N;if a + 1,S„ = a&#34;A +•for Vn e N.When a = 1, any particular solution is obtained by determining a specific,numerical value for I. In fact, a particular solution is determined by a specific,numerical value J for any (particular) entry, S,. Solving the equationJ = I+jc for I,1 = J – je.I/ since S; = I+jc// where So =1we get// One particular “particular solution&#34; has I = 0.When a + 1, any particular solution is obtained by determining a specific,numerical value for A; if the starting value I is given, then A = 1-,In fact,1- aa particular solution is determined by a specific, numerical value J for any(particular) entry, S,. Solving the equationJ = Ad +for A,awe getA =// But what if a = 0?// One particular “particular solution&#34; has A = 0.

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