Математикадан облыстық олимпиада, 1999-2000 оқу жылы, 9 сынып

Келесі теңдікті дәлелдеңіз: $\dfrac{\left( {{1}^{4}}+\dfrac{1}{4} \right)\left( {{3}^{4}}+\dfrac{1}{4} \right)\left( {{5}^{4}}+\dfrac{1}{4} \right)\ldots \left( {{11}^{4}}+\dfrac{1}{4} \right)}{\left( {{2}^{4}}+\dfrac{1}{4} \right)\left( {{4}^{4}}+\dfrac{1}{4} \right)\left( {{6}^{4}}+\dfrac{1}{4} \right)\ldots \left( {{12}^{4}}+\dfrac{1}{4} \right)}=\dfrac{1}{313}.$

Комментарий/решение

$$\prod \limits_{i=1}^{6}{\frac{(2k-1)^4+\frac{1}{4}}{(2k)^4+\frac{1}{4}}}=\frac{1}{313}$$

$$=\prod \limits_{i=1}^{6}{\frac{(2k-1)^4+\frac{1}{4}}{(2k)^4+\frac{1}{4}}}=\prod \limits_{i=1}^{6}{\frac{(2k-1)^4+2\cdot \frac{1}{2}\cdot (2k-1)^2+\frac{1}{4}-(2k-1)^2}{(2k)^4+2\cdot \frac{1}{2}\cdot (2k)^2+\frac{1}{4}-(2k)^4}}=$$

$$=\prod \limits_{i=1}^{6}{\frac{\left((2k-1)^2+\frac{1}{2}\right)^2-(2k-1)^2}{\left((2k)^2+\frac{1}{2}\right)^2-(2k)^2}}=$$

$$=\prod \limits_{i=1}^{6}{\frac{\left((2k-1)^2-(2k-1)+\frac{1}{2}\right) \cdot \left((2k-1)^2+(2k-1)+\frac{1}{2}\right)}{\left((2k)^2-2k+\frac{1}{2}\right) \cdot \left((2k)^2+2k+\frac{1}{2}\right)}}=$$

$$=\prod \limits_{i=1}^{6}{\frac{\left((2k-1)(2k-2) +\frac{1}{2}\right) \cdot \left((2k-1)2k+\frac{1}{2}\right)}{ \left((2k-1)2k+\frac{1}{2}\right) \cdot \left( 2k(2k+1)+\frac{1}{2}\right) }}=$$

$$=\prod \limits_{i=1}^{6}{\frac{\left((2k-1)(2k-2) +\frac{1}{2}\right)}{\left((2k+1)2k+\frac{1}{2}\right)}}=$$

$$=\frac{\frac{1}{2}}{\left(2\cdot3+\frac{1}{2}\right)}\cdot\frac{\left(2\cdot3+\frac{1}{2}\right)}{\left(5\cdot4+\frac{1}{2}\right)}\cdot\frac{\left(5\cdot4+\frac{1}{2}\right)}{\left(7\cdot6+\frac{1}{2}\right)}\cdot ... \cdot\frac{\left(9\cdot8+\frac{1}{2}\right)}{\left(11\cdot10+\frac{1}{2}\right)}\cdot\cdot\frac{\left(11\cdot10+\frac{1}{2}\right)}{\left(13\cdot12+\frac{1}{2}\right)}=$$

$$=\frac{\frac{1}{2}}{\left(13\cdot12+\frac{1}{2}\right)}=\frac{\frac{1}{2}}{\frac{313}{2}}=\frac{1}{313}$$