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Luly15
a fost răspuns

Demonstrati ca a1xa2xa3x...xan=[tex]
\sqrt{ (a1Xan) ^{n}
[/tex] 
a1 a2 a3 sunt termenii unei progreii geometrice 

Răspuns :

Matepentrutoti
Am atasat rezolvarea.
Vezi imaginea Matepentrutoti
[tex]\it a_1\cdot a_2\cdot a_3 \cdot ... \cdot a_n =a_1\cdot (a_1\cdot q) \cdot (a_1\cdot q^2)\cdot (a_1\cdot q^3)\cdot ... \cdot (a_1\cdot q^{n-1}) = \\\;\\ a_1^n\cdot q^{1+2+3+\ ...\ +n-1} =a_1^n\cdot q^{\dfrac{(n-1)n}{2}} = \left(a_1\cdot q^{\dfrac{n-1}{2}}\right)^n= \\\;\\ \\\;\\ =\left[(a_1^2)^{\frac{1}{2}} \cdot q^{\frac{n-1}{2} \right]^n = (a_1^2\cdot q^{n-1})^{n\cdot \frac{1}{2}} =(a_1\cdot a_1\cdot q^{n-1})^{n\cdot \frac{1}{2}} = \\\;\\ \\\;\\ = (a_1\cdot a_n)^{n\cdot \frac{1}{2}} =[/tex]

= [tex]\it \sqrt{(a_1\cdot a_n)^n}[/tex]


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