Question

(1 + 5 + 9 + 13 + ⋯ + 2017 + 2021) − (3 + 7 + 11 + ⋯ + 2015 + 2019)

Answer 1

Это выражение можно представить в виде двух арифметических прогрессий ($S_n$=$\frac{a1+a_n}{2}n$) нужно посчитать число n 1) d=4 4n-4=2021-1; 4n=2024 n=506; 2)d=4: 4n-4=2019-3; 4n=2020; n=505.

Считаем: $S1_5_0_6(=\frac{1+2021}{2}506) - S2_5_0_5 (= \frac{3+2019}{2} 505) = 511566 - 510555 = 1011$

Answer 2

It looks like you're trying to find the difference between two arithmetic series. The first series is an arithmetic sequence where each term increases by 4, starting from 1, and the last term is 2021. The second series is also an arithmetic sequence where each term increases by 4, starting from 3, and the last term is 2019.

To find the sum of an arithmetic series, you can use the formula:

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

where n is the number of terms in the series.

Let's first find the number of terms in each series:

For the first series: First term = 1 Last term = 2021 Common difference = 4

n = (last term - first term) / common difference + 1 n = (2021 - 1) / 4 + 1 n = 505

For the second series: First term = 3 Last term = 2019 Common difference = 4

n = (last term - first term) / common difference + 1 n = (2019 - 3) / 4 + 1 n = 504

Now, let's find the sum of each series using the formula:

Sum of the first series = (505/2) * (1 + 2021) = 510 * 2022 = 1030020 Sum of the second series = (504/2) * (3 + 2019) = 504 * 2022 = 1018848

Finally, the difference between the two sums is: Difference = Sum of the first series - Sum of the second series Difference = 1030020 - 1018848 = 11172

So, the result of the expression is 11172.