You haven’t provided a series so only a general answer can be given.
Assuming you have a finite series to work on you need to identify a pattern which gives you the next term in the series. If n is the position of a term in the series, n=1 would be the first term, n=2 the second term, and so on. Tn symbolises the nth term and Sn the sum of the terms up to the nth term.
The pattern you identify should help you to work out a formula for Tn as a function of n. For example, take the series: 8, 11, 14, 17, 20, ... You can guess the next term is 23 because there’s a constant difference of 3 between each term. So we have a function which gives us Tn, and we’ll call it T(n), so T(1)=8, T(2)=11, etc. If we were to plot the function we would have points (1,8), (2,11), (3,14), ... and they would connect with a straight line. The x values for these points are the n values for the position of each term. The formula for a straight line is y=mx+a where m is the slope of the line and a is some number we have to find. We can write this as T(n)=mn+a and T(n+1)=m(n+1)+a. So if we take two consecutive terms in the series T(1) and T(2), for example, we can work out T(2)-T(1), where n=1 and n+1=2. So T(2)-T(1)=11-8=3=2m+a-(m+a)=m, so m=3. Therefore, T(n)=3n+a. If we have n=1, then we know T(1)=8=3×1+a, that is, a+3=8, and a=8-3=5, so T(n)=3n+5. This is the formula to find each term. When n=6, T(6)=3×6+5=18+5=23, which was our guess for the next term. Of course, I used a very simple example, and many series are more complicated than this, but the principle is the same. Use observation and logic to find the formula.
But we also need to find Sn. That’s more involved...
We can find a formula for Sn and call it function S(n) just like we created the function T(n). We can relate two consecutive sums. The sum of the first n terms is S(n). The nth term is T(n). The sum to the previous term is S(n-1), for n>1, so S(n)=S(n-1)+T(n), in other words, we just add the last term to the sum of the previous terms to get a total sum of all n terms. We already have a formula for T(n)=3n+5 in our example, so S(n)=S(n-1)+3n+5.
But how do we find S(n-1)?
One way to do this is to write out every term from the first in longhand:
3×1+5 + 3×2+5 + 3×3+5 + ... +3n+5.
We can rearrange this sum:
3×1 + 3×2 + 3×3 + ... + 3n + 5 + 5 + 5 + ... + 5.
How many fives? There are n fives so that’s 5n.
And if we distribute 3 we get:
3(1+2+3+...+n) + 5n.
There’s a formula for finding the sum of the natural numbers from 1 to n. It’s n(n+1)/2.
So S(n)=3n(n+1)/2+5n=(3n²+3n+10n)/2=(3n²+13n)/2.
We can prove this formula by working out S(n-1) and adding on T(n):
S(n)=S(n-1)+T(n)=3(n-1)n/2+5(n-1) + 3n+5.
S(n)=(3n²-3n)/2+5n-5+3n+5=(3n²-3n)/2+8n=(3n²-3n+16n)/2,
S(n)=(3n²+13n)/2, which is the formula we got before.