Suppose that $gcd(a,b)=d$ and $gcd(a+b,b)=g$.
Then $d$ divides $a$ and $d$ divides $b$
Then $d$ divides $(a+b)$.
Again $d$ divides $b$.
Therefore $d$ divides $gcd(a+b,b)$,or in other words $d$ divides $g.$ ..................1
Next, $g$ divides $(a+b)$ and $g$ divides $b$.
Therefore $g$ divides $(a+b)-b$ or,$g$ divides $a$.
Again $g$ divides $b$.
Therefore $g$ divides $gcd(a,b)$ or in other words $g$ divides $d.$ ....................2
From 1 and 2 we can claim that $d$ divides $g$ as well as $g$ divides $d.$
This implies that $g=d.$
Therefore we have already proved the fact for t=1.
Now assume that the fact is true for t=n where n is a natural number.
Then by induction hypothesis $gcd(a,b)$= $gcd(an+b,b)$.
Again $gcd(an+b,b)$=$gcd(b+(n+1)a,b)$
Then $gcd(a,b)$= $gcd(an+b,b)$=$gcd(b+(n+1)a,b)$ Hence $gcd(a,b)$=$gcd(b+(n+1)a,b)$,and our proof is complete.