Question 1175082
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18 

<pre>
(a)  Find 1 + 2 + . . . + 24.

(b)  Show that 1/n, 2/n + ... n/n = (n+1)/2

(c)  Hence find the sum of the first 300 terms of
        1/1 + 1/2 + 2/2 + 1/3 + 2/3 + 3/3 + 1/4 + 2/4 + 3/4 +4/4 + ...
</pre>~~~~~~~~~~~~~~


<pre>
Well known fact is that the sum of the first n natural numbers

1 + 2 + 3 + . . . + n  is equal to  {{{(n*(n+1))/2}}}.     (1)


For the proof, see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/The-proofs-of-the-formulas-for-arithmetic-progressions.lesson>The proofs of the formulas for arithmetic progressions</A> 

in this site.



(a)  Therefore,  1 + 2 + 3 + . . . + 24 = {{{(24*25)/2}}} = 300.



(b)  From the formula (1),


         {{{1/n}}} + {{{2/n}}} + . . .  + {{{n/n}}} = {{{(1/n)*(n(n+1)/2)}}} = {{{(n+1)/2}}}.



(c)  Group the sum in this way


         Sum = (1/1) + (1/2 + 2/2) + (1/3 + 2/3 + 3/3) + (1/4 + 2/4 + 3/4 + 4/4) + . . .       (2)


     We have 300 terms/addends in all and the number of terms in k-th separate parentheses is k.


     Referring to the previous part (b) of this problem, we conclude that there are 24 groups in parentheses in the sum (2).


     Each particular group (k-th group) has the sum equal to {{{(k+1)/2}}}, according to part (b) of the solution.


     In other words,


         Sum = {{{sum((k+1)/2, k=1, 24)}}} = {{{(1/2)*(sum((k+1), k=1, 24))}}} = {{{(1/2)*(2 + 3 + ellipsis + 25)}}} = {{{(1/2)*((25*26)/2-1)}}} = {{{(1/2)*(650-1)}}} = {{{649/2}}} = 324 {{{1/2}}} = 324.5.    <U>ANSWER</U>
</pre>

Solved.


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<U>Post-solution note</U>


    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In this problem, &nbsp;its separate parts &nbsp;(a), &nbsp;(b) &nbsp;and &nbsp;(c) &nbsp;are logically inter-connected.