Question

1^2+3^2+5^2+...+(2n-1)^2

Answer 1

k cho t nhé

Answer 2

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Answer 3

có công thức: 1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6 (*)

đặt: S(2n + 1) = 1^2 + 3^2 + 5^2 + ... + (2n+1)^2

S(2n) = 2^2 + 4^2 + ...+ (2n)^2 = 2^2(1^2 + 2^2 + 3^2 +...+ n^2) = 4n(n+1)(2n+1)/6

=> S(2n+1) + S(2n) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + ... + (2n)^2 + (2n+1)^2

áp dụng (*) ta có: S(2n+1) + S(2n) = (2n+1)[(2n+1) + 1)[2(2n+1) + 1)]/6 = (2n+1)(2n+2)(4n+3)/6

=> S(2n+1) = (2n+1)(2n+2)(4n+3)/6 - S(2n) = 2(2n+1)(n+1)(4n+3)/6 - 4n(n+1)(2n+1)/6

= (2n+1)(n+1)/3.[(4n +3) - 2n] = (2n+1)(n+1)(2n+3)/3 vậy S(2n+1) = (2n+1)(n+1)(2n+3)/3

Answer 4

k cho t nhé