Solve sum with n and i

Question

I'm struggling to simplify this equation:

$$3 + \sum_{i=0}^{n-1} 18n-18i-32$$

Wolfram Alpha tells me that i can simplify this to $3+9n^2-23n$.

How ist that possible?

Thank you

Answer 1

$\displaystyle 3 + \sum_{i=0}^{n-1} (18n-18i-32)=3+18n^2-18\sum_{i=0}^{n-1}i-32n.$

Now, $\displaystyle \sum_{i=0}^{n-1}i=\frac{n(n-1)}{2}.$ Now it should be easy.

Answer 2

Since $18 n - 32$ is constant with respect to $i$, we have

$$\sum_{i = 0}^{n - 1} 18n - 32 = (18n - 32) \sum_{i = 0}^{n - 1} 1 = (18 n - 32) n$$

Finally, a general formula is that

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

Using this, the simplification is immediate.

Answer 3

1) $n (n-1)=n^2-n$

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

These are hints. Can you handle from here?