Arithmetic progression. find in terms of n

Question

Find in terms of n: $$\sum_{r=1}^{n} 2r-1$$ and $$ \sum_{r=0}^{n} 3r+3 $$

I tried using summation but they said in terms of n and then I expanded substituting from numbers 1 and 4.

Answer 1

Okay, lets use $$ \sum_{r=0}^{n} 3r+3 $$

as an example. I am going to break the two sums up into $$ \sum_{r=0}^{n} 3r+\sum_{r=0}^{n}3 $$ and notice that the term being summed over in the second sum does not change, it is simply $3+3+3+...+3$ added $n$ times.

That gives us $$3n+ \sum_{r=0}^{n} 3r $$

Focusing on $$\sum_{r=0}^{n} 3r $$

I'll pull out the $3$ leaving us with $$3\sum_{r=0}^{n} r $$

I am not sure if you have seen this but $$\sum_{r=0}^{n} r = \frac{n(n+1)}{2}$$

There is an explanation here: http://www.csun.edu/~ac53971/pump/20081014_basic.pdf

So, all totaled you have $$3\frac{n(n+1)}{2}+3n$$

But this can be used to solve any problem like that. For $$ \sum_{r=0}^{n} ar+b $$ you have $$a\frac{n(n+1)}{2}+bn$$