Предмет: Алгебра, автор: Kirofeed

Помогите решить! Задача для школьников, то есть метод математической индукции нужно как-то обойти.

Приложения:

BMW52: 470838
BMW52: Неверно.
Simba2017: может разбить на несколько сумм?

Ответы

Автор ответа: kamilmatematik100504
1

Ответ:

2·3 + 3·4 + ... + 98·99 = 323398

Объяснение:


Заметим что

2\cdot 3 + 3 \cdot 4 + 5 \cdot 6  \ldots  + 98 \cdot 99 =\\\\   6 + 12  + 30  + \ldots  + 98 \cdot 99   = \\\\ 3^2 - 3 + 4^2 - 4 + 5^2 - 5 + \ldots  + 99^2  -99 = \\\\ (3^2 + 4^2 + \dots  + 99^2 )  - ( 3 +  4 +  \ldots  + 99)

Выходит арифметическая  прогреcсия ,  а также сумма квадратов  чисел от 3 до 99

Для арифметической прогрессии все просто

S_{ap} =\dfrac{99 +3}{2}  \cdot 97  = 51 \cdot 97  = 4974

А    сумму квадратов  чисел от 3 до 99 , найдем с помощью формулы

S_{KB} = \dfrac{n\cdot (n+1)\cdot (2n+1)}{6}

( n  - количество чисел  от 1² + 2² ...  + n²)

S_{99} = \dfrac{99 \cdot (99+1)\cdot (99\cdot 2 +1)}{6 } = \dfrac{99 \cdot 100 \cdot 199}{6}  = 1650 \cdot 199 = \\\\ = 200 \cdot 1650 - 1650 = 330000 - 1650 =328350

Но наш ряд начинается с тройки , поэтому нужно отнять 1² + 2² = 5

S_{KB}  = 328350 - 5 = 328345

Теперь находим

(3^2 + 4^2 + \dots  + 99^2 )  - ( 3 +  4 +  \ldots  + 99) = \\\\S_{KB} - S_{ap} = 328345-4947    = 323398

#SPJ1





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