Correct option is AGiven: 4−51+13+12+144-\frac{5}{1+\frac{1}{3+\frac{1}{2 +\frac{1}{4}}}}4−1+3+2+41115 Solution: 4−51+13+12+14 =4−51+13+49 =4−51+931 =4−15540 =160−15540 =540 =184-\frac{5}{1+\frac{1}{3+\frac{1}{2 +\frac{1}{4}}}} \\ \ \\ = 4-\frac{5}{1+\frac{1}{3+\frac{4}{9}}} \\ \ \\ = 4-\frac{5}{1+\frac{9}{31}} \\ \ \\ = 4-\frac{155}{40} \\ \ \\ = \frac{160-155}{40} \\ \ \\ = \frac{5}{40} \\ \ \\ = \frac{1}{8} 4−1+3+2+41115 =4−1+3+9415 =4−1+3195 =4−40155 =40160−155 =405 =81