The method Charlie uses works by realising that in arithmetic series (arithmetic series are series where in order to get from one term to the next you add or subtract a fixed amount) the first and final term added together is the same as the second and second last term, etc.
So Charlie can simply add the first and last term, add these together, and then multiply this by the number of terms he has to start off with. This will give him what the sum of twice his original series was, as he "paired up" each term, with its corresponding term starting from the back. So he just divides his result by two.
This can be used for any arithmetic series, where the difference between terms is one, as long as you know the first and last term. The number of terms in these cases is given by:
( Last Term - First Term ) + 1.
1+2+3+...+19+20 = ?
1+20 = 21
(21*20)/2 = 210
( 1+2+3+...+99+100 = ?
1+100 = 101
(101*100)/2 = 5050
40+41+...+99+100 = ?
40+100 = 140
(140*61)/2 = 4270 )
In order to make it work for series' where the common difference is greater than 1, we simply need to find the number of terms in these cases. If we're given the first and last term, as well as common difference we can use:
[ ( Last Term - First Term ) / Difference ] + 1 . (We'll see why later *)
So:
1+3+5+...+19 = ?
Difference = 3-1 = 2.
Number of Terms = [(19 - 1)/2]+1 = 10
Sum: [(19+1)*10]/2 = 100
( 2+4+...+20 = ?
Difference = 4-2 = 2
Number of Terms = [(20-2)/2]+1 = 10
Sum: [(20+2)*10]/2 = 110
42+44+...+98+100 = ?
Difference = 44-42 = 2
Number of Terms = [(100-42)/2] +1 = 30
Sum = [(100+42)*30]/2 = 2130 )
We can use the above method to get an expression for 1+2+3+...+(n-1)+n
Difference = 2-1 = 1
Number of terms = [(n-1)/1]+1 = n
Sum = [(n+1)*n]/2 = (n^2 + n)/2
To understand why this works we need to set out some definitions:
In an arithmetic sequence there is a starting term, we'll call it a or U(1).
In an arithmetic sequence, to get from one term to another we simply add a common difference (d) to the previous term. U(65) = U(64) + d.
Using the above we see that U(1) = a, U(2) = a + d , U(3) = a + 2d . Following this trend we get U(n) = a + (n-1)d. Where n represents the term number. So the sum of the arithmetic series from U(1) to U(N) = U(1) + U(2) + ... + U(n).
Now if we use take this twice, and match the first term of our original with the last term of our second copy, and the second term with the second last term, etc. we get :
U(1)+U(n)+U(2)+U(n-1)+....+U(n)+U(1) = 2S(n) [ S(n) = Sum of first n terms ]
now putting in our definitions for the terms:
{a + [a + (n-1)d]}+ {a+d + [a + (n-2)d]} + ... + {[a + (n-1)d] + a} = 2S(n)
adding like terms withing the {} gives:
{2a+(n-1)d}+{2a+(n-1)d}+...+{2a+(n-1)d} = 2S(n)
We have {2a+(n-1)d} 'n' times, so we can rewrite it as:
n{2a+(n-1)d} = 2S(n)
Now we can divide by two to get S(n):
n/2{2a+(n-1)d} = S(n).
Now the {} was {a + [a + (n-1)d]}, which is equivalent to [U(1)+ U(n)]
So we have n/2[U(1)+U(n)] = S(n), which is the same as Charlie's method!
Now we can find the sum of any arithmetic series, like a Gauss.
* The number of terms can be found by using the equation U(n) = a + (n-1)d, this rearranges to: {[U(n) - a]/d} + 1 = n which is:
[(Last Term - First Term) / difference] + 1