Question

Comparați cu 1 fractia ordinară​

Answer 1

Răspuns:

f < 1

Explicație pas cu pas:

f = (2⁰+2¹+2²+............2⁴⁹) / (5⁰+5¹+5²+..........5²⁵) = a/b

a = 2⁰+2¹+2²+............2⁴⁹ I· 2  =>

2a = 2¹+2²+2³+......+2⁴⁹+2⁵⁰

2a = a-1 +2⁵⁰ => a = 2⁵⁰-1

b = 5⁰+5¹+5²+..........5²⁵  I·5 =>

5b = 5¹+5²+5³+.....+5²⁵+5²⁶

5b = b-1+5²⁶ => 4b = 5²⁶-1 =>

b = (5²⁶-1)/4

f = 4(2⁵⁰-1)/(5²⁶-1)

f < 1

2⁵⁰ = 2¹⁰ˣ⁵  = (2¹⁰)⁵ = 1024⁵

5²⁶ = 5·5⁵ˣ⁵ = 5·(5⁵)⁵ = 5·3125⁵ =>

4(2⁵⁰-1) < 5²⁶-1 => f < 1

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