Respuestas
Respuesta dada por:
2
Te ayudo con el segundo:
Primero debes saber algunas cosas:
![1+x+x^2+x^3+...+x^{n-1}= \frac{x^n-1}{x-1} \\ \\ x^{2n}-1=(x^n+1)(x^n-1) 1+x+x^2+x^3+...+x^{n-1}= \frac{x^n-1}{x-1} \\ \\ x^{2n}-1=(x^n+1)(x^n-1)](https://tex.z-dn.net/?f=1%2Bx%2Bx%5E2%2Bx%5E3%2B...%2Bx%5E%7Bn-1%7D%3D+%5Cfrac%7Bx%5En-1%7D%7Bx-1%7D+%5C%5C+%5C%5C+x%5E%7B2n%7D-1%3D%28x%5En%2B1%29%28x%5En-1%29)
Entonces en tu problema:
![1+2+2^2+2^3+...+2^{127}= \frac{2^{128}-1}{2-1}=2^{128}-1\\=(2^{64}+1)(2^{64}-1)\\=(2^{64}+1)(2^{32}+1)(2^{32}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^{16}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^8-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2+1)(2-1)\\ 1+2+2^2+2^3+...+2^{127}= \frac{2^{128}-1}{2-1}=2^{128}-1\\=(2^{64}+1)(2^{64}-1)\\=(2^{64}+1)(2^{32}+1)(2^{32}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^{16}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^8-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2+1)(2-1)\\](https://tex.z-dn.net/?f=1%2B2%2B2%5E2%2B2%5E3%2B...%2B2%5E%7B127%7D%3D+%5Cfrac%7B2%5E%7B128%7D-1%7D%7B2-1%7D%3D2%5E%7B128%7D-1%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B64%7D-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B32%7D-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E%7B16%7D-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E8-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E4%2B1%29%282%5E4-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E4%2B1%29%282%5E2%2B1%29%282%5E2-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E4%2B1%29%282%5E4-1%29%5C%5C%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E4%2B1%29%282%5E2%2B1%29%282%2B1%29%282-1%29%5C%5C)
![=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(3) =(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(3)](https://tex.z-dn.net/?f=%3D%282%5E%7B64%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B16%7D%2B1%29%282%5E8%2B1%29%282%5E4%2B1%29%282%5E2%2B1%29%283%29)
Entonces:
![\sqrt[32]{1+3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}\\= \sqrt[32]{1+2^{128}-1}= \sqrt[32]{2^{128}}=2^{128/32}=2^4=16 \sqrt[32]{1+3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}\\= \sqrt[32]{1+2^{128}-1}= \sqrt[32]{2^{128}}=2^{128/32}=2^4=16](https://tex.z-dn.net/?f=+%5Csqrt%5B32%5D%7B1%2B3%282%5E2%2B1%29%282%5E4%2B1%29%282%5E8%2B1%29%282%5E%7B16%7D%2B1%29%282%5E%7B32%7D%2B1%29%282%5E%7B64%7D%2B1%29%7D%5C%5C%3D+%5Csqrt%5B32%5D%7B1%2B2%5E%7B128%7D-1%7D%3D+%5Csqrt%5B32%5D%7B2%5E%7B128%7D%7D%3D2%5E%7B128%2F32%7D%3D2%5E4%3D16+++)
Saludos!
Primero debes saber algunas cosas:
Entonces en tu problema:
Entonces:
Saludos!
PascualDavid:
Ah y en el primero debes llegar a que x=y
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