Question

Să se demonstreze:

1/log2(1)+log2(2)+…+log2(2015)+
1/log3(1)+log3(2)+…+log3(2015)+…+
1/log2015(1)+log2015(2)+…+ log2015 (2015)=1

Answer 1

Răspuns:

1

Explicație pas cu pas:

$\boxed { log_{a}(x) + log_{a}(y) = log_{a}(xy) }$

$\boxed { \frac{1}{log_{a}(b)} = log_{b}(a) }$

$\boxed {log_{a}(a) = 1}$

$\frac{1}{ log_{2}(1) + log_{2}(2) + ... + log_{2}(2015) } + \frac{1}{ log_{3}(1) + log_{3}(2) + ... + log_{3}(2015) } + ... + \frac{1}{ log_{2015}(1) + log_{2015}(2) + ... + log_{2015}(2015) } = \frac{1}{ log_{2}(1 \cdot 2 \cdot ... \cdot 2015)} + \frac{1}{ log_{3}(1 \cdot 2 \cdot ... \cdot 2015)} + ... + \frac{1}{ log_{2015}(1 \cdot 2 \cdot ... \cdot 2015)} = log_{(1 \cdot 2 \cdot ... \cdot 2015)}(2) + log_{(1 \cdot 2 \cdot ... \cdot 2015)}(3) + ... + log_{(1 \cdot 2 \cdot ... \cdot 2015)}(2015) = log_{(2 \cdot ... \cdot 2015)}(2 \cdot ... \cdot 2015) = \bf 1$

q.e.d.