1. We are given $A=\{1,2,3,\dots,100\}$ and the relation $R=\{(a,b): a=2b+1\}.$

So every ordered pair in $R$ has the form $(2b+1,b)$ for some $b\in A$, with also $2b+1\in A$.

2. Since $2b+1\le 100$, we get $2b+1\le 100 \implies 2b\le 99 \implies b\le 49.5.$ Thus $b\le 49.$ Hence all pairs in $R$ are $(2b+1,b), \quad b=1,2,3,\dots,49.$

3. We need a sequence $(a_1,a_2),(a_2,a_3),(a_3,a_4),\dots,(a_k,a_{k+1})$ of elements of $R$.

Because each pair belongs to $R$, we must have $a_i=2a_{i+1}+1$ for each $i=1,2,\dots,k$.

So the chain looks like $a_1=2a_2+1,$ $a_2=2a_3+1,$ $a_3=2a_4+1, \dots$

4. To maximize $k$, we should start with the largest possible $a_1\le 100$ and keep generating the next terms by $a_{i+1}=\frac{a_i-1}{2}.$ For this to stay in integers, each current term must be odd.

So we want the longest chain obtained by repeatedly applying $n\mapsto \frac{n-1}{2}$ while staying in $A$.

5. Let us try the largest possible odd starting value, $99$: $99 \to 49 \to 24.$ But now $24$ is even, so it cannot be the first component of any pair in $R$ because numbers of the form $2b+1$ are odd. Thus this gives only $(99,49), (49,24),$ so $k=2$ here.

6. To get a long chain, every intermediate term $a_1,a_2,\dots,a_k$ must be odd, since each must be a first entry of a pair in $R$.

Using $a_i=2a_{i+1}+1,$ we can express $a_1$ in terms of $a_{k+1}$: $a_1=2^k a_{k+1} + (2^k-1).$ This is obtained by repeated substitution.

So $a_1+1 = 2^k(a_{k+1}+1).$ Since $a_1\le 100$ and $a_{k+1}\ge 1$, the smallest possible value of $a_1$ for a given $k$ occurs when $a_{k+1}=1$. Then $a_1=2^k\cdot 1 + (2^k-1)=2^{k+1}-1.$ For this to lie in $A$, $2^{k+1}-1\le 100.$ So $2^{k+1}\le 101.$ Now,