I think it can be simpler. Almost verbally. Let's examine a sequence where each subsequent term is twice the previous one, starting with 2. In essence, this will be a sequence where each subsequent term is 2 to the power of 1 greater than the previous one. That is: 2, 4, 8, 16,... = 2¹, 2², 2³, 2⁴,... . The sum of n terms of such a sequence will be (2ⁿ×2)-2. Thus, the sum of any number of terms of this sequence will always be in the range from the largest of them to what should be the next minus 2. This means that the sum must necessarily include 2 with the largest degree close to this number. In this case, 512, or 2^9. Another number: 640 - 512 = 128 =2^7. And since by the condition of the problem m
Que fantástica resolução.
Eu realmente adorei. ❤❤
Vendo as tuas resoluções dos exercícios de matemática, aí eu noto que eu estudo pouco. 😢😢
Thank you so much sir
2^m+2^n=640 m = 7, n = 9 m = 9, n = 7 It’s in my head.
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m,n (7,9)
Excellent
I think it can be simpler. Almost verbally.
Let's examine a sequence where each subsequent term is twice the previous one, starting with 2. In essence, this will be a sequence where each subsequent term is 2 to the power of 1 greater than the previous one. That is:
2, 4, 8, 16,... = 2¹, 2², 2³, 2⁴,... . The sum of n terms of such a sequence will be (2ⁿ×2)-2. Thus, the sum of any number of terms of this sequence will always be in the range from the largest of them to what should be the next minus 2. This means that the sum must necessarily include 2 with the largest degree close to this number. In this case, 512, or 2^9. Another number: 640 - 512 = 128 =2^7. And since by the condition of the problem m
Thank you so much
640=10×64=5×128=(1+4)128
=128+512=2^7+2^9
m=7,n=9
👌👌
@MathswithChinwendu thank you.