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i) Find the first five terms of the recursively defined sequence ak = 2ak-1 + k for all integers k ≥ 2 with a1 = 1. ii) Find a recurrence for the number tn of ternary words with no consecutive 0s. Then use this recurrence to find t6. iii) Solve the recurrence relation an = 5an-1 - 6an-2 with the initial conditions a0 = 1, a1 = 6. Prove using induction.

          i) Find the first five terms of the recursively defined sequence ak = 2ak-1 + k for all integers k ≥ 2 with a1 = 1.
ii) Find a recurrence for the number tn of ternary words with no consecutive 0s. Then use this recurrence to find t6.
iii) Solve the recurrence relation an = 5an-1 - 6an-2 with the initial conditions a0 = 1, a1 = 6. Prove using induction.
        
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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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i) Find the first five terms of the recursively defined sequence ak = 2ak-1 + k for all integers k ≥ 2 with a1 = 1. ii) Find a recurrence for the number tn of ternary words with no consecutive 0s. Then use this recurrence to find t6. iii) Solve the recurrence relation an = 5an-1 - 6an-2 with the initial conditions a0 = 1, a1 = 6. Prove using induction.
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Transcript

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00:01 Hello everyone.
00:01 So it is given that ak is equal to 2, ak minus 1 plus k.
00:06 Now, a1 is equal to 1.
00:10 It is given now putting k is equal to 2 3, 4, we will get a2 is 2, a2 minus 1 plus 2, that is equal to 2 plus 2 is equal to 4, a3 is equal to 2.
00:32 A3 minus 1 plus 3, that is 2 multiplied by 4 plus 3 that is equal to 11.
00:39 A4 is equal to 2a4 minus 1 plus 4 that is 2 multiplied by 11 plus 4 that is equal to 26 and a5 is equal to 2...
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