Question

Обчислити (1-1024)(2-1024)...(2014-1024)(2015-1024)

Answer 1

Ответ будет 0,т.к :
(1-1024)(2-1024)......(1024-1042)....
А мы знаем,что если умножить число на 0 , то ответ будет 0.

Answer 2

To calculate the product of the given expression (1-1024)(2-1024)...(2014-1024)(2015-1024), we can simplify it as follows:

(1-1024)(2-1024)(3-1024)...(2014-1024)(2015-1024)

Let's notice that each term in the parentheses is of the form (n - 1024), where n takes values from 1 to 2015. We can factor out -1024 from each term:

= (-1024)(1)(2)(3)...(2014)(2015)

Now, we need to calculate the product of the integers from 1 to 2015. This is known as the factorial of 2015, denoted as 2015!.

2015! = 1 × 2 × 3 × ... × 2013 × 2014 × 2015

However, calculating the factorial of such a large number can result in a very large value. Instead, we can use an approximation using Stirling's approximation:

n! ≈ √(2πn) * (n / e)^n

where e is the base of the natural logarithm (approximately equal to 2.71828).

Now, we can approximate 2015!:

2015! ≈ √(2π * 2015) * (2015 / e)^2015

Let's calculate this value step by step:

1. √(2π * 2015) ≈ √(4030π) ≈ 112.88
2. (2015 / e)^2015 ≈ (2015 / 2.71828)^2015

You can use a calculator to get this approximation:

(2015 / 2.71828)^2015 ≈ 3.51619 × 10^1163

Now, the approximate value of the product is:

≈ (-1024) × 112.88 × (3.51619 × 10^1163)

Finally, we can express the answer in scientific notation:

≈ -3.68785 × 10^1166

So, the approximate value of (1-1024)(2-1024)...(2014-1024)(2015-1024) is -3.68785 × 10^1166.