Page 5 - Matematica_Mathematics - Suma lui Gauss_Gauss sum

Practice

1. Following the solved model discovered by Gauss, calculate:
• 1+2+3+4+.....+23+24+25=

• 1+2+3+4+.....+47+48+48=

And because Gauss’ method is becoming more clear, we can go to the next level and calculate the sum
of the first n natural numbers. We apply the formula discovered by Gauss and say that:
1 + 2 + 3 + ........ + n = n × (n+1) : 2

We assume that we have to calculate the amount 6 + 9 + 12 + 15 + ..... + 198 + 201. We notice
that each term of the string is 3 times bigger than the previous one. We will write the sum twice, as Gauss did.

6 + 9 + 12 + 15 + ......... + 198 + 201 = S
201 + 198 + 195+ 192 + ......... + 9 + 6 = S

207 + 207 + 207+ 207 + ......... + 207 + 207 = 2 × S
We have one more obstacle - to find out how many terms this string has. Because the string is from 3 to
3, we will write the terms as products when a factor is 3, that is 2 × 3 + 3 × 3 + 4 × 3 + ............. + 66 × 3

+ 67 × 3. From 2 to 67 there are 66 numbers. So, adding up the terms by 2, we get 207 by 66 times. Basically,
our drill is almost done.
6 + 9 + 12 + 15 + ..... + 198 + 201 = 207 × 66 : 2

What result did you get by doing this calculation?

Practice
2. Calculate the sum:

• 5+10+15+....+60+65+70=