Can you help me simplify this summation notation?

Question

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

and

$$\sum_{i=1}^n \frac{i}{n}$$

(n is a constant)

Answer 1

The terms with $n$ are independent from the summation variable - which is $i$ - and can therefore be drawn in front of the summation sign. Thus $$\sum_{i=1}^{n}\frac{n}{n+1}i^2=\frac{n}{n+1}\sum_{i=1}^{n}i^2=\frac{n}{n+1}\frac{n(n+1)(2n+1)}{6}=\frac{n^2}{}$$ and $$\sum_{i=1}^{n}\frac{i}{n}=\frac{1}{n}\sum_{i=1}^{n}i=\frac{1}{n}\frac{n(n+1)}{2}$$ The formulas for summation of consequtive integers (or square of consequtive integers) can be found here.

Answer 2

There are well-known formulas for these sums:

$$\sum_{i=1}^n i=\frac{n(n+1)}{2}$$ and $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$$

You can prove them by recurrence.