1

This really is a question about the semantics of the sigma notation for writing long sums in a concise way. Let's say we have a sum given by the following notation,

$$\sum_{i = 0}^{n} (\frac{1}{n} + i^2)$$

My question now is rather simple. For a given $n$, let's say $n = 3$, would this sum written out look like this,

$$\sum_{i = 0}^{3} (\frac{1}{3} + i^2) = (\frac{1}{3} + 0^2) + (\frac{1}{3} + 1^2) + (\frac{1}{3} + 2^2) + (\frac{1}{3} + 3^2)$$

or like this,

$$\sum_{i = 0}^{3} (\frac{1}{3} + i^2) = \frac{1}{3} + 0^2 + 1^2 + 2^2 + 3^2$$

Basically, my question is, whether the constant term $\frac{1}{n}$ is only counted once in the sum or for every iteration. I realize this may seem like a stupid or obvious question to some, which is why its so difficult finding anything related to this specific thing.

roblox99
  • 427

1 Answers1

4

As summation is linear, we can actually think of the summation of a sum as a sum of summations: $$\sum\left(a+b\right)=\sum a+\sum b.$$

For your summation, we would be able to write $$\sum_{i=0}^n\left(\frac1n+i^2\right)=\sum_{i=0}^n\frac1n+\sum_{i=0}^ni^2.$$

As such, for $n=3$, we have $$\sum_{i=0}^3\frac13+\sum_{i=0}^3i^2=\underbrace{\left(\frac13+\frac13+\frac13+\frac13\right)}_{\text{one for each index number}}+\left(0^2+1^2+2^2+3^2\right)=\frac43+14=\frac{46}3.$$

Andrew Chin
  • 7,389