Question #96920
Find the H.M. of 1/2,2/3,3/4,.......n/(n+1) occurring with frequencies 1,2,3,.....n
1
Expert's answer
2019-10-21T10:42:05-0400

The harmonic mean can be found as


HM=k=1nfkk=1nfkxkHM = {{\sum\limits_{k = 1}^n {{f_k}} } \over {\sum\limits_{k = 1}^n {{{{f_k}} \over {{x_k}}}} }}

where xk{{x_k}} are dara values and fk{{f_k}} are corresponding frequencies. We've got fk=k{f_k} = k and xk=kk+1{x_k} = {k \over {k + 1}} . Thus we have to find two sums. The first is just the arithmetic progression


k=1nfk=k=1nk=n(n+1)2\sum\limits_{k = 1}^n {{f_k}} = \sum\limits_{k = 1}^n k = {{n(n + 1)} \over 2}

The second can be simplifid to the arithmetic progression and the sum of units


k=1nfkxk=k=1nkkk+1=k=1n(k+1)=k=1nk+k=1n1=n(n+1)2+n\sum\limits_{k = 1}^n {{{{f_k}} \over {{x_k}}}} = \sum\limits_{k = 1}^n {{k \over {{k \over {k + 1}}}}} = \sum\limits_{k = 1}^n {(k + 1)} = \sum\limits_{k = 1}^n k + \sum\limits_{k = 1}^n 1 = {{n(n + 1)} \over 2} + n

Then


HM=n(n+1)2n(n+1)2+n=2n(n+1)+2nn(n+1)2=n+1n+3HM = {{{{n(n + 1)} \over 2}} \over {{{n(n + 1)} \over 2} + n}} = {2 \over {n(n + 1) + 2n}} \cdot {{n(n + 1)} \over 2} = {{n + 1} \over {n + 3}}


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