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1+12(53)+34(63)+(73)+54(83)+32(93)+74(103)=(ab) 1 + \frac{1}{2}{5\choose3}+\frac{3}{4}{6\choose3}+{7\choose3}+\frac{5}{4}{8\choose3}+\frac{3}{2}{9\choose3}+\frac{7}{4}{10\choose3} = {a\choose b}1+21(35)+43(36)+(37)+45(38)+23(39)+47(310)=(ba)
It is given that 2b≤a2b \le a2b≤a and aaa is as small an integer as possible. Determine the value of a+ba+ba+b.
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