Question

Vreau si eu va rog rezolvarea la exercițiu! mersi ​

Answer 1

$\sqrt{ \frac{1 + 3 + 5 + ... + 47}{1 + 3 + 5 + ... + 23} } \\ \\ = \sqrt{ \frac{1 + (1 + 2) + (1 + 4) + ... + (1 + 46)}{1 + (1 + 2) + (1 + 4) + ... +(1 + 22) } } \\ \\ = \sqrt{ \frac{\underbrace{1 + 1 + ... + 1}_{24} + 2 + 4 + ... + 46}{\underbrace{1 + 1 + ... + 1} _{12} + 2 + 4 + ... + 22} } \\ \\ = \sqrt{ \frac{24 + 2(1 + 2 + 3 + .. + 23)}{12 + 2(1 + 2 + .. + 11)} } \\ \\ = \sqrt{ \frac{24 + 2 \times \frac{24 \times 24}{2} }{12 + 2 \times \frac{11 \times 12}{2} } } \\ \\ = \sqrt{ \frac{24 + 2 \times 23 \times 12}{12 + 2 \times 11 \times 6} } \\ \\ = \sqrt{ \frac{24 + 552}{12 + 132} } \\ \\ = \sqrt{ \frac{576}{144} } \\ \\ = \frac{24}{12} \\ \\ = 2$

Answer 2

Explicație pas cu pas:

sumă Gauss pentru numere impare:

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

a)

$\sqrt{ \frac{1 + 3 + 5 + ... + 47}{1 + 3 + 5 + ... + 23} } = \sqrt{ \frac{ {24}^{2} }{ {12}^{2} } } = \sqrt{ {\Big( \frac{24}{12} \Big)}^{2} } = \sqrt{ {2}^{2} } = \bf 2$