Ответ:
4074341
Объяснение:
Пусть 2017 = k ; k(k+1)(k+2)(k+3) + 1 = (k·(k+3))·((k+1)·(k+2)) + 1 =
(k²+3k)·(k²+3k+2)+1 = (k²+3k)² + 2(k²+3k) + 1 = (k²+3k +1)² = n²
при k= 2017 n = 4074341
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Verified answer
Ответ:
4074341
Объяснение:
Пусть 2017 = k ; k(k+1)(k+2)(k+3) + 1 = (k·(k+3))·((k+1)·(k+2)) + 1 =
(k²+3k)·(k²+3k+2)+1 = (k²+3k)² + 2(k²+3k) + 1 = (k²+3k +1)² = n²
при k= 2017 n = 4074341