[tex]I_n= \int\limits^1_2 {(x-1)^n(x-2)^n} \, dx =\int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}(x^2-3x+2) \, dx (1)
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[tex]I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-1)^{n+1}]'(x-2)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}(x-1)^2} \, dx (2)[/tex]
[tex]I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-2)^{n+1}]'(x-1)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-2)^{n-1}(x-1)^{n-1}(x-2)^2} \, dx(3)[/tex]
Se observa ca:[tex]2(x^2-3x+2)-(x-1)^2-(x-2)^2=-1\\
Din\ 2(1)+ \frac{n+1}{n} (2)+\frac{n+1}{n} (3) \ obtinem:\\
2I_n+ \frac{n+1}{n}I_n+ \frac{n+1}{n}I_n= \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}\cdot (-1)} \, dx=-I_{n-1}\\
I_n \frac{4n+2}{n} =-I_{n-1}\\
(4n+2)I_n+nI_{n-1}=0[/tex]
