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
14
1
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.
(xn+2-1)/(x2-1)
its 12(s-1/2)(s+1) is that what you were looking for?
-1
s=4
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
It is 35.
ANSWER:1+1= 2/1=2+1=3/1=3+1=4/1=4+1=5/1=5+1=6the answer is 6answer is 2.If you put parentheses after all your "divided by" 's and go on forever, this converges to the golden ratio.
That factors to (m + n)(r + s) The GCF is 1.
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Add all the 1's except the last one, since 1 x 0 = 0. If you want to do all the additions first (this should be indicated by proper placing of parentheses), then the result is zero, since anything times zero = zero. In this case, you don't actually have to carry out the addition.