Question

A12. Să se calculeze sumele:
subpunctul d mă interesează ​

Answer 1

Explicație pas cu pas:

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

$\boxed {1 + 3+ 5 + ... + (2n - 1) = {n}^{2}}$

$\boxed {1 + 2 + 3 + ... +n = \frac{n(n + 1)}{2}}$

A.12.d)

$log_{a}{x}^{} \sqrt[n]{ {x}^{2} } + log_{a}{x}^{3} \sqrt[n]{ {x}^{4} } + ... + log_{a}{x}^{2n - 1} \sqrt[n]{ {x}^{2n} } =$

$= log_{a}(x \cdot {x}^{ \frac{2}{n} }) + log_{a}( {x}^{3} \cdot {x}^{ \frac{4}{n} }) + ... + log_{a}( {x}^{2n - 1} \cdot {x}^{ \frac{2n}{n} })$

$= log_{a}({x}^{} \cdot {x}^{ \frac{2}{n} } \cdot {x}^{3} \cdot {x}^{ \frac{4}{n} } \cdot ... \cdot {x}^{2n - 1} \cdot {x}^{ \frac{2n}{n} })$

$= log_{a}({x}^{1 + 3 + ... + (2n - 1)} \cdot {x}^{ \frac{2 + 4 + ... + 2n}{n}})$

$= log_{a}({x}^{ {n}^{2} } \cdot {x}^{ \frac{2 \cdot (1 + 2 + ... + n)}{n}}) = log_{a}({x}^{ {n}^{2} } \cdot {x}^{ \frac{2 \cdot \frac{n(n + 1)}{2} }{n}})$

$= log_{a}({x}^{ {n}^{2} } \cdot {x}^{n + 1}) = log_{a}({x}^{ {n}^{2} + n + 1})$