$S=\frac{1^{4}}{1\times 3}+\frac{2^{4}}{3\times 5}+....+\frac{n^{4}}{(2n-1)(2n+1)}$
Đặt $S_n=\frac{1^{4}}{1.3}+\frac{2^{4}}{3.5}+....+\frac{n^{4}}{(2n-1)(2n+1)}$
Ta chứng minh $S_n=\frac{n(n+1)(n^2+n+1)}{6(2n+1)}, \forall n\ge 1$ 
*) Với $n=1$, ta có: $S_1=\frac{1}{3}=\frac{1.2.3}{6.3}$, đúng.
*) Giả sử mệnh đề đúng với $n=k,k\ge 1$, tức $ S_k=\frac{k(k+1)(k^2+k+1)}{6(2k+1)} $ 
Ta chứng minh mệnh đề đúng với $n=k+1$ 
Ta có:
$S_{k+1}=S_k+\frac{(k+1)^4}{(2k+1)(2k+3)}$ 
            $= \frac{k(k+1)(k^2+k+1)}{6(2k+1)}+\frac{(k+1)^4}{(2k+1)(2k+3)} $ 
            $=\frac{(k+1)[k(k^2+k+1)(2k+3)+6(k+1)^3]}{6(2k+1)(2k+3)}$ 
            $=\frac{(k+1)(2k^4+11k^3+23k^2+21k+6)}{6(2k+1)(2k+3)}$  
            $=\frac{(k+1)(k+2)(2k+1)(k^2+3k+3)}{6(2k+1)(2k+3)}$ 
            $=\frac{(k+1)(k+2)[(k+1)^2+(k+1)+1]}{6(2k+3)}$ , đpcm.
Vậy: $S_n=\frac{n(n+1)(n^2+n+1)}{6(2n+1)}, \forall n\ge 1$  
thank b nha –  tanbinhtho 15-10-12 07:09 PM

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