(a) Find a closed-form solution for this recurrence relation: an=2⋅an−1−n+1 with a1=2a (b) Prove that your closed-form solution is correct.

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Fatema Sutton · Accepted Answer

(a) Given an=(2an−1)−n+1 a1=2 Let us determine the first terms of the sequence until we notice a pettern: a1=2=1+1 a2=2a1−2+1=2(2)−2=4−1=3=2+1 a3=2a2−3+1=2(3)−3=6−1=4=3+1 a4=2a3−4+1=2(4)−4=8−1=5=4+1 a5=2a4−5+1=2(5)−5=10−1=6=5+1 a6=2a5−6+1=2(6)−6=12−1=7=6+1⋯ Thus we note that an=n+1 appears to hold in general. (b) Given an=(2an−1)−n+1 a1=2 To proof: an=n+1 for all positive integers. Proof by induction Let P(n) be the statement an=n+1 Basis step n=1 an=a1=2 n+1=1+1=2 Thus P(1) is true Inductive step Let P(k) be true. ak=k+1 We need to proof that P(k+1) is true. ak+1=2a((k+1)−1)−(k+1)+1=(2ak)−k  =2(k+1)−k=2k+2−k=k+2=(k+1)+1 Thus P(k+1) is true. Conclusion By the principle of mathematical induction, P(n) is true for all positive integers n.

(a) Find a closed-form solution for this recurrence relation: an=2⋅an−1−n+1 with a1=2a (b) Prove that your closed-form solution is correct.

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