(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.

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.