Question

Нужна помощь с объяснением решения данного задания! Задание было: Вычислить сумму первых k слагаемых ряда : 1+1/(2^2)-1/(3^2)+1/(4^2)-...1/(k^2) Вот решение. var k,z,i:integer; s:real; begin readln(k); z:=-1; s:=1; for i:=2 to k do begin z:=-z; s:=s+z/(i*i); end; writeln('s = ',s); end. КАК тут было учтено то, что когда делим на нечетное число в квадрате, то знак стоит "-"? В каких строчках? И как это объяснить?

Answer 1

Смотрите в строчке под номером 4 
z:=-1; z становится отрицательным
Потом его пустили по циклу, в первом круге цикла этот же z (z:=z;) становится с противоположным знаком(z:=-(-1);) Тоесть сначала выполняется действие добавления(+), в следующем кругу уже положительному z меняют знак на противоположный, и z опять у нас
с "-" и так k раз.

Вроде понятно обьяснил, ну если что обращайтесь.

Answer 2

Explanation of the Solution

The given code calculates the sum of the first k terms of a series. The series is defined as follows:

1 + 1/(2^2) - 1/(3^2) + 1/(4^2) - ... + (-1)^(k+1) / (k^2)

Let's go through the code step by step to understand how it handles the alternating signs in the series.

Code Explanation

var k, z, i: integer;
    s: real;
begin
    readln(k);
    z := -1;
    s := 1;
    for i := 2 to k do
    begin
        z := -z;
        s := s + z / (i * i);
    end;
    writeln('s += ', s);
end.

1. The code declares four variables: `k`, `z`, `i`, and `s`. - `k` is used to store the value of `k` entered by the user. - `z` is used to alternate the sign of each term in the series. It is initialized to -1. - `i` is a loop variable used in the `for` loop. - `s` is used to store the sum of the series.

2. The code reads the value of `k` from the user.

3. The code initializes `z` to -1 and `s` to 1. This is because the first term of the series is always 1.

4. The code enters a `for` loop that iterates from 2 to `k`. This loop calculates the sum of the remaining terms in the series.

5. Inside the loop, the code alternates the sign of `z` by multiplying it by -1. This is done using the statement `z := -z`. When `z` is -1, it becomes 1, and vice versa.

6. The code calculates the value of the current term in the series by dividing `z` by `(i * i)`. The term is then added to the sum `s`.

7. After the loop finishes, the code outputs the value of the sum `s`.

Handling the Alternating Signs

The alternating signs in the series are handled by the variable `z`.

- Initially, `z` is set to -1. This means that the first term of the series, which is always 1, will be added to the sum `s`. - Inside the `for` loop, the sign of `z` is alternated by multiplying it by -1 using the statement `z := -z`. This means that for each iteration of the loop, the sign of `z` will change from positive to negative or from negative to positive. - The current term in the series is then calculated by dividing `z` by `(i * i)`. Since `z` alternates between positive and negative, the sign of the term will also alternate accordingly.

Let's take an example to understand this better. Suppose `k` is 5.

1 + 1/(2^2) - 1/(3^2) + 1/(4^2) - 1/(5^2)

- The first term is always 1, so it is added to the sum `s`. - In the first iteration of the loop, `z` is -1. The term `1/(2^2)` is negative, so it is subtracted from the sum `s`. - In the second iteration, `z` becomes 1. The term `1/(3^2)` is positive, so it is added to the sum `s`. - In the third iteration, `z` becomes -1 again. The term `1/(4^2)` is negative, so it is subtracted from the sum `s`. - In the fourth iteration, `z` becomes 1 again. The term `1/(5^2)` is positive, so it is added to the sum `s`.

The final value of `s` will be the sum of these terms.

Conclusion

In conclusion, the code handles the alternating signs in the series by using a variable `z` that alternates between -1 and 1. The sign of each term in the series is determined by the value of `z`. When `z` is -1, the term is subtracted from the sum `s`, and when `z` is 1, the term is added to the sum `s`. This allows the code to correctly calculate the sum of the first k terms of the series.