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Show that the sum of the end term of the progression Log x, Log x^2, Log x^3, log x^4 = n (n+1/2) Log x
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Wording and writing in the post are incorrect, so I edited it to make sense from nonsense.
My edited formulation is as follows:
Show that the sum of n terms of the progression log(x), log(x^2), log(x^3), log(x^4) , . . , log(x^n) is (n*(n+1)/2)*log x.
Below is my solution for this edited formulation.
In this problem, x > 0.
Let a = log(x).
Then = 2a, = 3a, . . . and so on . . . till = n*a.
Therefore, our progression takes the form
a + 2a + 3a + . . . + n*a = (1 + 2 + 3 + . . . + n)*a.
The sum in parentheses in the right side is well known sum of the arithmetic progression,
and it is equal to .
Thus + + + . . . + = .
It is the final answer, and the proof is complete.
Solved.