udowodnij MM: Udowodnij, że jeżeli b=a−1, to (a+b)(a²+b²)(a⁴+b⁴)(a⁸+b⁸)...(a³²+b³²)=a⁶⁴−b⁶⁴

think: a⁶⁴ − b⁶⁴ = {ze wzoru skróconego mnożenia a² − b² = (a + b)(a − b)} = (a³² + b³²)(a³² − b³²) = {nadal z tego samego wzoru} (a³² + b³²)(a¹⁶ + b¹⁶)(a¹⁶ − b¹⁶) = (a³² + b³²)(a¹⁶ + b¹⁶)(a⁸ + b⁸)(a⁸ − b⁸) = (a³² + b³²)(a¹⁶ + b¹⁶)(a⁸ + b⁸)(a⁴ + b⁴)(a⁴ − b⁴) = (a³² + b³²)(a¹⁶ + b¹⁶)(a⁸ + b⁸)(a⁴ + b⁴)(a² + b²)(a² − b²) = (a³² + b³²)(a¹⁶ + b¹⁶)(a⁸ + b⁸)(a⁴ + b⁴)(a² + b²)(a + b)(a − b) = (a³² + b³²)(a¹⁶ + b¹⁶)(a⁸ + b⁸)(a⁴ + b⁴)(a² + b²)(a + b)(a − (a − 1)) = cdn.

MM: dzięki wielkie oczywiście