Question 1209833: Fill in the blanks, to make a true equation.
3/(3^2 - 1) + 3^2/(3^4 - 1) + 3^3/(3^6 - 1) + 3^4/(3^8 - 1) + ... + 3^(2(n - 1))/(3^(2n) - 1) = ___/___
Answer by ikleyn(52775)   (Show Source):
You can put this solution on YOUR website! .
Fill in the blanks, to make a true equation.
3/(3^2 - 1) + 3^2/(3^4 - 1) + 3^3/(3^6 - 1) + 3^4/(3^8 - 1) + ... + 3^(2(n - 1))/(3^(2n) - 1) = ___/___
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As I read your post attentively, I see that the general expression for the last, n-th term
3^(2(n - 1))/(3^(2n) - 1)
is not consistent with the expressions for first and third terms.
Indeed, for the first term n= 1, so the first term, according to the general formula, should be
3^(2(1 - 1))/(3^(2*1) - 1) = 3^0/(3^2-1) = 1/(3^2-1),
but in your long formula the first term is different.
Next, for the third term n= 3, so the third term, according to the general formula, should be
3^(2(3 - 1))/(3^(2*3) - 1) = 3^4/(3^6-1),
but in your long formula the third term is different.
Thus, again, the general formula for the n-th term is not consistent with the formulas for the 1st and 3rd terms.
This makes the problem to look like as a lame horse (with three legs).
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