Help with the following succession I tried this problem a few times, but couldn’t really solve it anyway this is the problem:

Determine the sum of the following $51$ numbers of the succession
   $a_n=3+2n$ for all whole numbers $n>0$

Can someone help me solve this ? Thanks 
 A: Hints:
$\sum\limits_{n=1}^{51} 
\left(3+2n\right) = 3\left(\sum\limits_{n=1}^{51} 1\right) + 2\left(\sum\limits_{n=1}^{51} n\right)$
Then, recognize the sums that remain.  In particular, recognize and remember what you know about triangle numbers.
A: You will learn as you progress with this topic that the sum of an arithmetic progression $S_n$ is $$S_n=\frac{n}{2}(2a_1+(n-1)d),$$ where $a_i$ is the $i^{\text{th}}$ term in the sequence and $d$ is the common difference.
So here, we have $n=51$, $a_1=3+2(1)=5$ and $d=2$, so $$S_{51}=\frac{51}{2}(2(5)+(50\times2))=\frac{51}{2}(111)=2805.$$
A: The answer to your question is that you would do the following:

$$\sum_\limits{n=1}^{51}(3+2n)$$

From there you would use the following rules:
Summation Rules
The addition/subtraction of summations rule:
$$\sum_\limits{i=1}^a(X_i\pm Y_i)=\sum_\limits{i=1}^a(X_i)\pm \sum_\limits{i=1}^a(Y_i)$$
Allowing you to breakdown the problem into the following method:

$$\sum_\limits{n=1}^{51}(3)+\sum_\limits{n=1}^{51}(2n)$$

Then using another two rules
Rule 1:
$$\sum_\limits{n=1}^{a}(c)=ca$$
Rule 2:
$$c\sum_\limits{n=1}^{a}(Y_i)$$
Thus giving you the following:

$$51(3)+2\sum_\limits{n=1}^{51}(n)$$

Another handy rule used in summations is the following which is:
$$\sum_\limits{n=1}^{a}(n)=\frac{a(a+1)}2$$
Which then gives you after cleaning up

$$153+2(\frac{51(51+1)}2)$$

Which is equal to $2085$.
