let 1+2+4+8+...... =s s= 1+2(1+2+4+8+........) s=1+2s s=-1 but sum of infinite terms cannot be -1 so, what's the answer
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Another Answer: 12*0+1 = 1
x3 + 1 = (x + 1)(x2 - x + 1) The x + 1's cancel out, leaving x2 - x + 1
let 1+2+4+8+...... =s s= 1+2(1+2+4+8+........) s=1+2s s=-1 but sum of infinite terms cannot be -1 so, what's the answer
1S+1S+1 = 3 Note that 1 raised to any power is always 1.
no it;s not the answer is 2
(xn+2-1)/(x2-1)
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its 12(s-1/2)(s+1) is that what you were looking for?
That depends on the value of s.
s=4
s = 2
7 is the answer. The order of operations would be to multiply the 2 times 1 and then add the five other 1's. (Unless there are parenthesis around some of the numbers) Here are the orders # Parentheses # Exponents # Multiplication and Division # Addition and Subtraction
you could try "what will 1 1 212 1 2 3 4 n one =?" and add it up like this.:):):):):):):) or i don't know... 1 1 1 2 3 4 +212=224
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