+0

0
414
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Guest Jan 17, 2015

#2
+81022
+10

Very nice, geno.....!!!

I'm not as crafty as geno, so I had to "brute-force" this one to see if I could discover another possible pattern....!!!

(1+2) + (1+3) + (1+4) + (1+5) + (1+6) + (1+7)

+  (2+3) + (2+4) + (2+5) + (2+6) + (2+7)

+ (3+4) + (3+5) + (3+6) + (3+7)

+ (4+5) + (4+6) + (4+7)

+ (5+6) + (5+7)

+ (6+7)

-------------------------------------------------------------------

(1+2)+(3+6)+(6+12)+(10+20)+(15+30)+(21+42) =

Notice that the sum of the first terms in the parentheses is just 56 = (8 x 7) = 1(8 x 7)

And notice that the sum of the second terms in the parentheses is just  112   = 2(8 x 7)

So we have 1(8 x 7)+ 2(8 x 7) = 3(8 x 7)= 8(7 x 3) = 8 x 21  =  (7+1)C(7,2) = (8)C(7,2)

Notice that the "pattern," is just (n+1)C(n,2)  ....where n = 7

BTW....this result can be extended to any two element subset sums of the first n positive integers.....for instance.....the sum of the two element subsets of the first 10 integers =(10+1)C(10,2)  = 11(45) = 495 ......

----------------------------------------------------------------------------------------------------

See the inductive proof of this below........

CPhill  Jan 17, 2015
Sort:

#1
+17711
+10

When creating these two elements subsets, you must use each of the elements 6 times (so each can be matched with each of the other 6 numbers).

So you need 6 1's, 6 2's, 6 3's, ..., 6 7's.

6[1 + 2 + 3 + 4 + 5 + 6 + 7]  =  6[28]  =  168

geno3141  Jan 17, 2015
#2
+81022
+10

Very nice, geno.....!!!

I'm not as crafty as geno, so I had to "brute-force" this one to see if I could discover another possible pattern....!!!

(1+2) + (1+3) + (1+4) + (1+5) + (1+6) + (1+7)

+  (2+3) + (2+4) + (2+5) + (2+6) + (2+7)

+ (3+4) + (3+5) + (3+6) + (3+7)

+ (4+5) + (4+6) + (4+7)

+ (5+6) + (5+7)

+ (6+7)

-------------------------------------------------------------------

(1+2)+(3+6)+(6+12)+(10+20)+(15+30)+(21+42) =

Notice that the sum of the first terms in the parentheses is just 56 = (8 x 7) = 1(8 x 7)

And notice that the sum of the second terms in the parentheses is just  112   = 2(8 x 7)

So we have 1(8 x 7)+ 2(8 x 7) = 3(8 x 7)= 8(7 x 3) = 8 x 21  =  (7+1)C(7,2) = (8)C(7,2)

Notice that the "pattern," is just (n+1)C(n,2)  ....where n = 7

BTW....this result can be extended to any two element subset sums of the first n positive integers.....for instance.....the sum of the two element subsets of the first 10 integers =(10+1)C(10,2)  = 11(45) = 495 ......

----------------------------------------------------------------------------------------------------

See the inductive proof of this below........

CPhill  Jan 17, 2015
#3
+81022
+5

Inductive proof that the total sum of all of the possible two element subset sums of the first n positive integers is given by .... (n + 1)C(n, 2)

Show that it is true for the first two positive integers

(2 + 1)C(2,2)  = 3(1) = 3

Asuume it's true for n = k....that is......

(k + 1)C(k, 2) = (k+1) [k!]/[ (k - 2)! 2!] = (k-1)(k)(k+1) / 2!

Prove it's true for n =  k + 1  ....that is...(k+1)C(k+1, 2) = (k)(k+1)(k+2) /2!

So we have

[(k + 1) + 1)] C(k+1, 2) =

(k + 1)C(k+1, 2)   + C(k+1, 2) =

(k + 1)[(k + 1)!] / [(k-1)!2!]  + [(k + 1)!] / [(k-1)!2!] =

(k + 1)[ (k+1)(k)] /2!  + [(k+1)(k) /2!] =

[(k + 1)(k) /2!] [ (k+1) + 1) ] =

[(k + 1)(k) /2!] [ (k+2) ] =

(k)(k+1)(k+2) / 2!   ......which is what we wished to show....

CPhill  Jan 19, 2015
#4
+18829
+5

John computes the sum of the elements of each of the 21 two-element subsets of {1,2,3,4,5,6,7}.

What is the sum of these 21 sums ?

$$\small{\text{  sum = \frac{3}{2} * \left(\ 1*2+2*3+3*4+4*5+5*6+6*7\ \right) = \frac{3}{2} * \left(\ 2+6+12+20+30+42\ \right)  }}\\ \small{\text{  = \frac{3}{2}*112=168  }}\\\\ \small{\text{ for n:  sum = \frac{3}{2} * \left(\ 1*2+2*3+3*4+4*5+5*6+6*7+\dots + (n-1)*n\ \right) = 3*\left( \begin{array}{c} n+1\\3 \end{array} \right)  }}\\\\ \small{\text{  = \dfrac{(n-1)n(n+1)}{2}  }}\\\\ \small{\text{ for n=7:  sum = 3*\left( \begin{array}{c} 8\\3 \end{array} \right) =3*\frac{6*7*8}{2*3}=21*8=168  }}$$

heureka  Jan 20, 2015

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