(a) Given \(an=(2an-1)-n+1\)

\(a1=2\)

Let us determine the first terms of the sequence until we notice a pettern:

\(a1=2=1+1\)

\(a2=2a1-2+1=2(2)-2=4-1=3=2+1\)

\(a3=2a2-3+1=2(3)-3=6-1=4=3+1\)

\(a4=2a3-4+1=2(4)-4=8-1=5=4+1\)

\(a5=2a4-5+1=2(5)-5=10-1=6=5+1\)

\(a6=2a5-6+1=2(6)-6=12-1=7=6+1\cdots\)

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.