Toán chứng minh đẳng thức

[linhntmk123]

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chứng minh đẳng thức sau :
[tex]1^{5}[/tex] +[tex]2^{5}[/tex] +[tex]3^{5}[/tex] +......+[tex]n^{5}[/tex]= [n^2. (n+2)^2. (2n^2 +2n-1)]/12

 

[Blue Plus]

chứng minh đẳng thức sau :
[tex]1^{5}[/tex] +[tex]2^{5}[/tex] +[tex]3^{5}[/tex] +......+[tex]n^{5}[/tex]= [n^2. (n+2)^2. (2n^2 +2n-1)]/12

Ta có:
$ n^5\\ = \frac{12n^5}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2 - 2n^6 + 6n^5 - 5n^4 + n^2}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2}{12} - \frac{2n^6 - 6n^5 + 5n^4 - n^2}{12} \\ = \frac{2n^6 + 2n^5 - n^4 + 4n^5 + 4n^4 - 2n^3 + 2n^4 + 2n^3 - n^2}{12} - \frac{2n^6 - 2n^5 - n^4 - 4n^5 + 4n^4 + 2n^3 + 2n^4 - 2n^3 - n^2}{12} \\ = \frac{n^4(2n^2 + 2n - 1) + 2n^3(2n^2 + 2n - 1) + n^2(2n^2 + 2n - 1)}{12} - \frac{n^4(2n^2 - 2n - 1) - 2n^3(2n^2 - 2n - 1) + n^2(2n^2 - 2n - 1)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^4 + 2n^3 + n^2)}{12} - \frac{(2n^2 - 2n - 1)(n^4 - 2n^3 + n^2)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^2 . n^2 + n^2 . 2n + n^2)}{12} - \frac{(2n^2 - 4n + 2 + 2n - 2 - 1)(n^2 . n^2 - n^2 . 2n + n^2)}{12} \\ = \frac{(2n^2 + 2n + 1)n^2(n^2 + 2n + 1)}{12} - \frac{[2(n^2 - 2n + 1) + 2(n - 1) - 1]n^2(n^2 - 2n + 1)}{12} \\ = \frac{(2n^2 + 2n - 1)n^2(n + 1)^2}{12} - \frac{[2(n - 1)^2 + 2(n - 1) - 1]n^2(n - 1)^2}{12} $
$ 1^5 = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} \\ 2^5 = \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} \\ 3^5 = \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} \\ ... \\ \Rightarrow 1^5 + 2^5 + 3^5 + ... + n^5 \\ = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} + \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} + \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} + ... + \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{(n- 1)^2 . n^2 . [2 . (n - 1)^2 + 2 .(n - 1) - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{0^2 . 1^2 . [2 . 0^2 + 2 . 0 - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - 0 \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12}(ĐPCM) $

 

[linhntmk123]

Ta có:
$ n^5\\ = \frac{12n^5}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2 - 2n^6 + 6n^5 - 5n^4 + n^2}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2}{12} - \frac{2n^6 - 6n^5 + 5n^4 - n^2}{12} \\ = \frac{2n^6 + 2n^5 - n^4 + 4n^5 + 4n^4 - 2n^3 + 2n^4 + 2n^3 - n^2}{12} - \frac{2n^6 - 2n^5 - n^4 - 4n^5 + 4n^4 + 2n^3 + 2n^4 - 2n^3 - n^2}{12} \\ = \frac{n^4(2n^2 + 2n - 1) + 2n^3(2n^2 + 2n - 1) + n^2(2n^2 + 2n - 1)}{12} - \frac{n^4(2n^2 - 2n - 1) - 2n^3(2n^2 - 2n - 1) + n^2(2n^2 - 2n - 1)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^4 + 2n^3 + n^2)}{12} - \frac{(2n^2 - 2n - 1)(n^4 - 2n^3 + n^2)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^2 . n^2 + n^2 . 2n + n^2)}{12} - \frac{(2n^2 - 4n + 2 + 2n - 2 - 1)(n^2 . n^2 - n^2 . 2n + n^2)}{12} \\ = \frac{(2n^2 + 2n + 1)n^2(n^2 + 2n + 1)}{12} - \frac{[2(n^2 - 2n + 1) + 2(n - 1) - 1]n^2(n^2 - 2n + 1)}{12} \\ = \frac{(2n^2 + 2n - 1)n^2(n + 1)^2}{12} - \frac{[2(n - 1)^2 + 2(n - 1) - 1]n^2(n - 1)^2}{12} $
$ 1^5 = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} \\ 2^5 = \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} \\ 3^5 = \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} \\ ... \\ \Rightarrow 1^5 + 2^5 + 3^5 + ... + n^5 \\ = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} + \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} + \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} + ... + \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{(n- 1)^2 . n^2 . [2 . (n - 1)^2 + 2 .(n - 1) - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{0^2 . 1^2 . [2 . 0^2 + 2 . 0 - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - 0 \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12}(ĐPCM) $

cảm ơn bn nhưng mà mk chứng minh dc rồi

 

[tôi là ai?]

Ta có:
$ n^5\\ = \frac{12n^5}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2 - 2n^6 + 6n^5 - 5n^4 + n^2}{12} \\ = \frac{2n^6 + 6n^5 + 5n^4 - n^2}{12} - \frac{2n^6 - 6n^5 + 5n^4 - n^2}{12} \\ = \frac{2n^6 + 2n^5 - n^4 + 4n^5 + 4n^4 - 2n^3 + 2n^4 + 2n^3 - n^2}{12} - \frac{2n^6 - 2n^5 - n^4 - 4n^5 + 4n^4 + 2n^3 + 2n^4 - 2n^3 - n^2}{12} \\ = \frac{n^4(2n^2 + 2n - 1) + 2n^3(2n^2 + 2n - 1) + n^2(2n^2 + 2n - 1)}{12} - \frac{n^4(2n^2 - 2n - 1) - 2n^3(2n^2 - 2n - 1) + n^2(2n^2 - 2n - 1)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^4 + 2n^3 + n^2)}{12} - \frac{(2n^2 - 2n - 1)(n^4 - 2n^3 + n^2)}{12} \\ = \frac{(2n^2 + 2n - 1)(n^2 . n^2 + n^2 . 2n + n^2)}{12} - \frac{(2n^2 - 4n + 2 + 2n - 2 - 1)(n^2 . n^2 - n^2 . 2n + n^2)}{12} \\ = \frac{(2n^2 + 2n + 1)n^2(n^2 + 2n + 1)}{12} - \frac{[2(n^2 - 2n + 1) + 2(n - 1) - 1]n^2(n^2 - 2n + 1)}{12} \\ = \frac{(2n^2 + 2n - 1)n^2(n + 1)^2}{12} - \frac{[2(n - 1)^2 + 2(n - 1) - 1]n^2(n - 1)^2}{12} $
$ 1^5 = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} \\ 2^5 = \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} \\ 3^5 = \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} \\ ... \\ \Rightarrow 1^5 + 2^5 + 3^5 + ... + n^5 \\ = \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} - \frac{0^2 . 1^2 . (2 . 0^2 + 2 . 0 - 1)}{12} + \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} - \frac{1^2 . 2^2 . (2 . 1^2 + 2 . 1 - 1)}{12} + \frac{3^2 . 4^2 . (2 . 3^2 + 2 . 3 - 1)}{12} - \frac{2^2 . 3^2 . (2 . 2^2 + 2 . 2 - 1)}{12} + ... + \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{(n- 1)^2 . n^2 . [2 . (n - 1)^2 + 2 .(n - 1) - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - \frac{0^2 . 1^2 . [2 . 0^2 + 2 . 0 - 1]}{12} \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12} - 0 \\ = \frac{n^2 . (n + 1)^2 . (2 . n^2 + 2 . n - 1)}{12}(ĐPCM) $

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