(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.

(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.

Question
Sequences
asked 2021-01-31
(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.

Answers (1)

2021-02-01
(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.
PSKak+1=2a((k+1)-1)-(k+1)+1 =(2ak)-k =2(k+1)-k =2k+2-k =k+2 =(k+1)+1ZSK
Thus P(k+1) is true.
Conclusion By the principle of mathematical induction, P(n) is true for all positive integers n.
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