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 Sequence of wrong means (Posted on 2011-05-26)
A) Start a sequence with any two real numbers, for the third and successive terms take the mean of all of the previous terms.

B) Same as part A) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one less than the number of terms.

C) Same as part A) and B) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one more than the number of terms.

What happens to the sequence in each case? Generalize further.

 See The Solution Submitted by Jer No Rating

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 induction does it | Comment 2 of 3 |

A) Mean(n)=1/n*(a+b) <br><br>

B) Mean(n)=(a+b)<br><br>

C) Mean(n)=1/3*(a+b) <br><br>

All the a/m answers can be easily obtained by induction. e.g.

case B: Mean(n+1)= (n*(Mean(n) +a+b)/(n+1)= a+b   q.e.d. <br>

 Posted by Ady TZIDON on 2011-05-27 05:46:38

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