How do you calculate the sum of all the terms of a succession?

The German Mathematician Carl Friedrich Gauss is said, when he was still a very young student, to have surprised his teacher when he suggested his students to calculate the value resulting from the sum of the whole numbers from 1 to 100. After a while Gauss would have answered, after using the sum strategy, 100 + 1, 99 + 2, 98 + 3, 97 + 4 and so on. He managed to obtain 100 times the sum 101. It means that if you add the numbers from 1 to 100 twice, you will obtain the result 10100. Thus, the answer to the question suggested by the professor is 5050 (the half of 10100).

Is there a formula for that?

Of course there is a formula and that’s what I want to hit by telling you the story about Gauss. The strategy he used stresses that in an arithmetic progression the sum of the equidistant terms is always the same. Thus, to get the sum of all the terms, it is enough to sum the first term with the last one, then you multiply that value by the amount of terms that you want to sum and finally you divide it by 2. Here it is the formula according to which n represents the amount of terms of the arithmetic progression that we want to sum: `S_n = (U_1 + U_n)/2 xx n`.

And... what happens with the sum of the terms of a geometric progression?

I am not going to tell you any story this time (because nobody knows if it is true or not). I just give you a formula according to which r stands for the ratio and the variable n represents the amount of terms of the geometric progression that we want to sum up: `S_n = U_1 xx (1-r^n)/(1-r)`.

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