Question 1210233
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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.



<pre>
In this problem, x > 0.


Let a = log(x).


Then {{{log((x^2))}}} = 2a,  {{{log((x^3))}}} = 3a, . . . and so on . . . till  {{{log((x^n))}}} = 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  {{{(n*(n+1))/2}}}.


Thus  {{{log((x))}}} + {{{log((x^2))}}} + {{{log((x^3))}}} + . . . + {{{log((x^n))}}} = {{{((n*(n+1))/2)*log((x))}}}.


It is the final answer, and the proof is complete.
</pre>

Solved.