Question

1+2+3+4+5+6+7+8+9...и так до 365

Answer 1

Ответ:

$S_{365}=66795$

Пошаговое объяснение:

дана конечная арифметическая прогрессия $1;2;3;4;5;...;365$, в которой $a_1=1,~a_n=365,~d=1$.

• найдем, сколько всего членов содержится в данной прогрессии по формуле n-го члена $\bf a_n=a_1+(n-1)*d$, для этого выразим n из формулы и подставим значения $a_1,a_n$ и $d$:  $a_n=a_1+(n-1)*d;\\a_n=a_1+nd-d;\\nd=a_n-a_1+d;\\n=\frac{a_n-a_1+d}{d} =\frac{365-\not1+\not1}{1} =365$
• вычислим $S_{365}$ по формуле суммы первых n членов арифметической прогрессии $\bf S_n=\frac{n*(a_1+a_n)}{2}$:  $S_{365}=\frac{365*(1+365)}{2} =\frac{365*\not366}{\not2} =365*183=\bf66795$

Answer 2

Sum of Numbers from 1 to 365

To find the sum of numbers from 1 to 365, we can use a formula for the sum of an arithmetic series. The formula is:

Sum = (n/2) * (first term + last term)

In this case, the first term is 1 and the last term is 365, and there are 365 terms in total. Plugging these values into the formula, we get:

Sum = (365/2) * (1 + 365) = 66795 [[9]](https://getcalc.com/math-sum-natural-numbers-1-365.htm).

Therefore, the sum of numbers from 1 to 365 is 66795.

Alternate Method

Another way to find the sum of numbers from 1 to 365 is to observe that the sequence forms an arithmetic progression with a common difference of 1. In this case, we can use the formula:

Sum = (n/2) * (first term + last term)

Using this formula, we get the same result:

Sum = (365/2) * (1 + 365) = 66795 [[8]](https://www.cuemath.com/questions/1-2-3-4-5-6-7-8-9/).

Conclusion

The sum of numbers from 1 to 365 is 66795.