John computes the sum of the elements of each of the 15 two-element subsets of $\{1,2,3,4,5,6\}$. What is the sum of these 15 sums?
It might be interesting to see if we can discover a pattern here.
Note
Three elements....two at time (1,2) (1,3) (2,3)....sum = 12
Four elements .... two at a time (1,2) (1,3) (1,4) (2,3) (2,4) (3,4) .....sum = 30
Five elements....two at a time (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5) ....sum = 60
Note that for N elements taken 2 at a time......the pattern of sums seems to be
(N - 1) ( N ) ( N+ 1) / 2
So....for 6 elements taken 2 at a time.....the sum should be
(5) (6) (7) / 2 = 210 / 2 = 105
Verify for yourself that this is true.......!!!!!!
As an alternative answer to this.....
Note that each element will be summed 5 times
So we have the sum (1 + 2 + 3 + 4 + 5 + 6) summed 5 times
And the sum of the frist 6 elements = (6) (7) / 2 = 21
So......the total sum will be
(5) * [ (6) * ( 7) / 2 ] = (N -1) (N) ( N + 1) / 2 which is the correct "formula"
John computes the sum of the elements of each of the 15 two-element subsets of
\(\{1,2,3,4,5,6\} \).
What is the sum of these 15 sums?
15 two-element subsets:
\(\begin{array}{|llllll|} \hline &1,& 2,& 3,& 4,& 5,& 6 \\ \hline &1,2 & 2,3 & 3,4 & 4,5 & 5,6 \\ &1,3 & 2,4 & 3,5 & 4,6 & \\ &1,4 & 2,5 & 3,6 & \\ &1,5 & 2,6 & \\ &1,6 \\ \hline \end{array}\)
\(\begin{array}{|lcll|} \hline \text{the numbers are: } && 5\times 1 \\ &+& 5\times 2 \\ &+& 5\times 3 \\ &+& 5\times 4 \\ &+& 5\times 5 \\ &+& 5\times 6 \\ \hline &=& 5\times (1+2+3+4+5+6) \\ &=& 5\times \left(\frac{1+6}{2}\right)\times 6 \\ &=& 5\times \left(\frac{7}{2}\right)\times 6 \\ &=& 105 \\\\ &=& (n-1)\times \left(\frac{1+n}{2}\right)\times n \\ &=& 3\times \binom{n+1}{3} \\ \hline \end{array} \)