sadiamek wrote:
Can anyone help please. Difficult to explain to my son.
We write S(2,5) as an abbreviation for 2+3+4+5 so that S(2,5) = 14.
Similarly, S(6,39) = 6+7+8+9+......+38 +39 = 765
Work out:
S(1,3)
S(6,40)
S(7,38)
S(1,2) - S(2,3) + S(3,4) - S(4,5) + ....... - S(18,19) + S(19,20)
Many thanks in advance
I think the last question is quite elegant.
S(1,2) - S(2,3) = 1 - 3
S(1,2) - S(2,3) + S(3,4) = 1+ 4
S(1,2) - S(2,3) + S(3,4) - S(4,5) = 1 - 5
Expanding similarly
S(1,2) - S(2,3) + S(3,4) ............ + S(19,20) = 1 + ?
Question like S(7,38) is quite brutal for a primary school child IMO.
Method suggested by moved of course works. To write down / count the number of pairs may not be that easy.
A secondary school kid could have tried to solved it using the formula for sum of first n natural numbers I,e, n(n+1)/2
S(7,38) therefore would have been 38*39/2 - 6*7/2. How a primary school child is supposed to know this though
.