Question 991818
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{{{1/log(2,a)}}} + {{{1/log(4,a)}}} + {{{1/log(8,a)}}} + . . . . . upto n term
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Notice that 


{{{1/log(b,a)}}} = {{{log(a,b)}}}.


It is the consequence of the &nbsp;"base change formula" &nbsp;for logarithms &nbsp;(see any &nbsp;<B>Algebra</B>&nbsp; textbook or the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/logarithm/change-of-base-formula-for-logarithms.lesson>Change of Base Formula for logarithms</A>&nbsp; in this site). 


Therefore, &nbsp;we can re-write the formula in the form


{{{1/log(2,a)}}} + {{{1/log(4,a)}}} + {{{1/log(8,a)}}} + . . . + {{{1/log(2^n,a)}}} = {{{log(a,2)}}} + {{{log(a,4)}}} + {{{log(a,8)}}} + . . . + {{{log(a,2^n)}}} 


Next, &nbsp;notice that &nbsp;&nbsp;{{{log(a,4)}}} = {{{2*log(a,2)}}}, &nbsp;&nbsp;{{{log(a,8)}}} = {{{3*log(a,2)}}}, . . . , &nbsp;&nbsp;{{{log(a,2^n)}}} = {{{n*log(a,2)}}}.


Thus you have, &nbsp;actually, &nbsp;the sum  


{{{log(a,2)}}} + {{{2*log(a,2)}}} + {{{3*log(a,2)}}} + . . . + {{{n*log(a,2)}}} = {{{log(a,2)}}}.(1 + 2 + 3 + . . . +n),


which is equal to &nbsp;&nbsp;{{{log(a,2)}}}.{{{((n(n+1))/2)}}}.


<U>Answer</U>. &nbsp;The sum is equal to &nbsp;{{{log(a,2)}}}.{{{((n(n+1))/2)}}} = {{{(1/log(2,a))}}}.{{{((n(n+1))/2)}}}.


Thank you for submitting so exciting problem.