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
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\ \
/ (2.i-1) or / f(i)
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i=1 i=1
In general: if f(i) is an expression depending on i
n
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\
/ f(i) = f(m) + f(m + 1) + f(m + 2) + ... + f(n)
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i=m
But, in this html document, for convenience, I'll write :
for i = m..n
sumi f(i) = f(m) + f(m + 1) + f(m + 2) + ... + f(n)
for i = 1..5
sumi (i.i + 1) = 2 + 5 + 10 + 17 + 26
or even :
for i = 1..5
sum (i.i + 1) = 2 + 5 + 10 + 17 + 26
If 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)
sumi ( sumj f(i,j) ) = sumj( sumi f(i,j) )
Therefore we can simply write the above sums as
sumi,j f(i,j)
sumi,j (f(i,j).g(j)) = sumj ( g(j).sumi f(i,j) )
sumi,j (f(i).g(j)) = sumi g(j) . sumj g(j)