Question

1+2+3+...+60= 2+4+6+...+40= 3+6+9+...+30= Приём глауса

Answer 1

1+60 =2+59 = 3 +58  = 4+57  итд и всего таких пар будет 30  следовательно нужно умножить на 30

30(1+60)=30 *61=1830 так же с другими попробуйте !

Только числа другие

Answer 2

It seems like you are asking about the sum of arithmetic progressions and the Gauss formula. Let's calculate the sums you mentioned:

1. Sum of the first 60 natural numbers: The sum of the first n natural numbers can be calculated using the formula Sn = n * (n + 1) / 2. For n = 60: S60 = 60 * (60 + 1) / 2 = 60 * 61 / 2 = 30 * 61 = 1830.

2. Sum of even numbers from 2 to 40: Since these numbers form an arithmetic progression with a common difference of 2, we can use the formula for the sum of an arithmetic progression: Sn = n * (a + l) / 2, where n is the number of terms, a is the first term, and l is the last term. a = 2, l = 40, and the common difference d = 2. n = (l - a) / d + 1 = (40 - 2) / 2 + 1 = 39 / 2 + 1 = 20.

S20 = 20 * (2 + 40) / 2 = 20 * 42 / 2 = 20 * 21 = 420.

3. Sum of multiples of 3 from 3 to 30: The first term (a) is 3, the last term (l) is 30, and the common difference (d) is also 3 (since we are dealing with multiples of 3). n = (l - a) / d + 1 = (30 - 3) / 3 + 1 = 27 / 3 + 1 = 9 + 1 = 10.

S10 = 10 * (3 + 30) / 2 = 10 * 33 / 2 = 330 / 2 = 165.

Finally, the formula you mentioned, known as the Gauss formula or the arithmetic series formula, helps calculate the sum of an arithmetic progression: Sn = n * (a + l) / 2, where Sn is the sum of the first n terms, a is the first term, l is the last term, and n is the number of terms.

So, to summarize: 1 + 2 + 3 + ... + 60 = 1830. 2 + 4 + 6 + ... + 40 = 420. 3 + 6 + 9 + ... + 30 = 165.