Question

find the value of (1⁰+2⁰+3‐¹+4‐¹)×(2-¹)²​

Answer 1

Answer

31/48

Solution

$({1}^{0} + {2}^{0} + {3 }^{ - 1} + {4}^{ - 1} ) \times {( {2}^{ - 1})}^{2} \\ \\ = (1 + 1 + \frac{1}{ {3}^{1} } + \frac{1}{ {4}^{1} } ) \times ( { \frac{1}{2} })^{2} \\ \\ = (2 + \frac{1}{3} + \frac{1}{4} ) \times \frac{1}{ {2}^{2} } \\ \\ = ( \frac{2 \times 12 + 4 + 3}{12} ) \times \frac{1}{4} \\ \\ = \frac{31}{12} \times \frac{1}{4} \\ \\ = \frac{31}{48}$

Answer 2

Answer:

$= (1 {}^{0} + {2}^{0} + {3}^{ - 1} + {4}^{ - 1} ) \times ( {2}^{ - 1} ) {}^{2} \\ = (1 + 1 + \frac{1}{3} + \frac{1}{4} ) \times ( \frac{1}{2} ) {}^{2} \\ = ( \frac{12 + 12 + 4 + 3}{12} ) \times ( \frac{1}{4} ) \\ = \frac{31}{12} \times \frac{1}{4} \\ = \frac{31}{48}$