Let mathrmA = \1, 2, 3, dots, 100\$\mathrm{A} = \{1, 2, 3, \dots, 100\}$ and mathrmR$\mathrm{R}$ be a relation on mathrmA$\mathrm{A}$ such that mathrmR = \(a, b) : a = 2b + 1\$\mathrm{R} = \{(a, b) : a = 2b + 1\}$. Let (a_1, a_2), (a_2, a_3), (a_3, a_4), dots, (a_k, a_k+1)$(a_1, a_2), (a_2, a_3), (a_3, a_4), \dots, (a_k, a_{k+1})$ be a sequence of k$k$ elements of mathrmR$\mathrm{R}$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k$k$, for which such a sequence exists, is equal to:

### Related Formula textChain definition: a_i = 2 a_i+1 + 1 quad textfor i = 1, 2, dots, k$$\text{Chain definition: } a_{i} = 2 a_{i+1} + 1 \quad \text{for } i = 1, 2, \dots, k$$ ### Core Logic To find the longest sequence of connected pairs, we trace the relation backward starting from the smallest elements in A$A$. ### Step 1: Trace the relations backward To maximize k$k$, we want the chain of elements to go down as low as possible. Let the final element in the chain be a_k+1 in mathrmA$a_{k+1} \in \mathrm{A}$. Since a_k = 2 a_k+1 + 1$a_k = 2 a_{k+1} + 1$: - If a_k+1 = 1 implies a_k = 3$a_{k+1} = 1 \implies a_k = 3$ - If a_k+1 = 2 implies a_k = 5$a_{k+1} = 2 \implies a_k = 5$ Let's test the chain starting with a_k+1 = 1$a_{k+1} = 1$: - a_k = 2(1) + 1 = 3$a_k = 2(1) + 1 = 3$ - a_k-1 = 2(3) + 1 = 7$a_{k-1} = 2(3) + 1 = 7$ - a_k-2 = 2(7) + 1 = 15$a_{k-2} = 2(7) + 1 = 15$ - a_k-3 = 2(15) + 1 = 31$a_{k-3} = 2(15) + 1 = 31$ - a_k-4 = 2(31) + 1 = 63$a_{k-4} = 2(31) + 1 = 63$ - a_k-5 = 2(63) + 1 = 127$a_{k-5} = 2(63) + 1 = 127$ (but 127 notin mathrmA$127 \notin \mathrm{A}$!) Thus, the longest chain within the set A$A$ has 6 elements: \63, \, 31, \, 15, \, 7, \, 3, \, 1\$$\{63, \, 31, \, 15, \, 7, \, 3, \, 1\}$$ This chain corresponds to exactly 5 ordered pairs: (63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$(63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$ So the maximum number of pairs in the sequence is k = 5$k = 5$. ### Step 2: Check alternative chains If we start with a_k+1 = 2$a_{k+1} = 2$: - a_k+1 = 2$a_{k+1} = 2$ - a_k = 5$a_k = 5$ - a_k-1 = 11$a_{k-1} = 11$ - a_k-2 = 23$a_{k-2} = 23$ - a_k-3 = 47$a_{k-3} = 47$ - a_k-4 = 95$a_{k-4} = 95$ - a_k-5 = 191 > 100$a_{k-5} = 191 > 100$ Again, the maximum number of pairs is k = 5$k = 5$. Thus, the largest integer k$k$ is 5. ### Pattern Recognition Recursive scaling: Tracing exponential chains of the form x_n+1 = c x_n + d$x_{n+1} = c x_n + d$ shows that the elements grow very quickly. Calculating the limits of growth determines the maximum possible depth of the sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions