Question

[rus] найдите N если 1 + 2 + 3 + 4 + ...+ N = 1275 [eng] Find N if 1 + 2 + 3 + 4 + ...+ N = 1275​

Answer 1

Ответ:

N=50  

Пошаговое объяснение:

Сумма этой арифметической прогрессии

равна N*(N+1)/2

N*(N+1)=2550

можно решить квадратное уравнение

N*N+N-2550=0 (найти корни 50 и -51 и отрицательный отбросить)

а можно заметить, что 2550=50*51

N=50

Answer 2

To find the value of N in the equation 1 + 2 + 3 + 4 + ... + N = 1275, you can use the formula for the sum of an arithmetic series:

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

In this case, the first term is 1, and the last term is N. We are given that the sum is 1275. Plugging these values into the formula:

1275 = (N/2) * (1 + N)

Now, let's solve for N:

1275 = (N/2) * (1 + N)

Multiply both sides by 2 to isolate N:

2 * 1275 = N * (1 + N)

2550 = N^2 + N

Rearrange the equation to a quadratic form:

N^2 + N - 2550 = 0

Now, you can solve this quadratic equation for N. You can use the quadratic formula:

N = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 1, b = 1, and c = -2550. Plugging in these values:

N = (-1 ± √(1² - 4(1)(-2550))) / (2(1))

N = (-1 ± √(1 + 10200)) / 2

N = (-1 ± √10201) / 2

N = (-1 ± 101) / 2

Now, you have two possible solutions:

1. N = (-1 + 101) / 2 = 100 / 2 = 50
2. N = (-1 - 101) / 2 = -102 / 2 = -51

Since N represents a positive number in this context, the solution is N = 50.

So, N = 50 is the value that satisfies the equation 1 + 2 + 3 + 4 + ... + N = 1275.