Let f(i) = (2.i-1) with i an integer .
Suppose that you want to calculate f(1) + f(2) + ... + f(32)
In math we denote this sum with a sigma sign as
32 32 --- --- \ \ / (2.i-1) or / f(i) --- --- i=1 i=1In general: if f(i) is an expression depending on i
n --- \ / f(i) = f(m) + f(m + 1) + f(m + 2) + ... + f(n) --- i=mBut, in this html document, for convenience, I'll write :
for i = m..n sum_{i} f(i) = f(m) + f(m + 1) + f(m + 2) + ... + f(n) for i = 1..5 sum_{i} (i.i + 1) = 2 + 5 + 10 + 17 + 26or even :
for i = 1..5 sum (i.i + 1) = 2 + 5 + 10 + 17 + 26If the limits for i have no importance, we'll write
sum f(i)
sum c.f(i) = c. sum f(i)
sum (f(i) + g(i)) = sum f(i) + sum g(i)
sum_{i} ( sum_{j} f(i,j) ) = sum_{j}( sum_{i} f(i,j) )Therefore we can simply write the above sums as
sum_{i,j} f(i,j)
sum_{i,j} (f(i,j).g(j)) = sum_{j} ( g(j).sum_{i} f(i,j) )
sum_{i,j} (f(i).g(j)) = sum_{i} g(j) . sum_{j} g(j)