10Toán 10 Trắc nghiệm Thông hiểu 1,936 Xét mệnh đề P(n)P(n)P(n): “12+22+32+⋯+n2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n\left(n+1\right)\left(2n+1\right)}{6}12+22+32+⋯+n2=6n(n+1)(2n+1)” với mọi số nguyên dương nnn, chứng minh bằng quy nạp toán học. Giả thiết quy nạp là A 12+22+32+⋯+k2=k(k+1)(2k+1)61^2 + 2^2 + 3^2 + \cdots + k^2 = \dfrac{k\left(k+1\right)\left(2k+1\right)}{6}12+22+32+⋯+k2=6k(k+1)(2k+1) B 12+22+32+⋯+k+12=k+1(k+1+1)(2k+1+1)61^2 + 2^2 + 3^2 + \cdots + k+1^2 = \dfrac{k+1\left(k+1+1\right)\left(2k+1+1\right)}{6}12+22+32+⋯+k+12=6k+1(k+1+1)(2k+1+1) C 12+22+32+⋯+n2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n\left(n+1\right)\left(2n+1\right)}{6}12+22+32+⋯+n2=6n(n+1)(2n+1) D 12+22+32+⋯+k2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + \cdots + k^2 = \dfrac{n\left(n+1\right)\left(2n+1\right)}{6}12+22+32+⋯+k2=6n(n+1)(2n+1) Câu hỏi có hữu ích? 28 0