Question

4．用二项式定理证明：3²ⁿ-8n-1能被64整除（n∈N^*）．

Answer 1

分析 由3²ⁿ-8n-1=（8+1）ⁿ-8n-1，利用二项式定理展开化简为64（ ${C}_{n}^{0}$•8^n-2+${C}_{n}^{1}$•8^n-3+${C}_{n}^{2}$•8^n-4+…+${C}_{n}^{n-2}$），从而证得结论．

解答 解：3²ⁿ-8n-1=（8+1）ⁿ-8n-1=${C}_{n}^{0}$•8ⁿ+${C}_{n}^{1}$•8^n-1+${C}_{n}^{2}$•8^n-2+…+${C}_{n}^{n-1}$•8+${C}_{n}^{n}$-8n-1
=${C}_{n}^{0}$•8ⁿ+${C}_{n}^{1}$•8^n-1+${C}_{n}^{2}$•8^n-2+…+${C}_{n}^{n-2}$•8²=64（ ${C}_{n}^{0}$•8^n-2+${C}_{n}^{1}$•8^n-3+${C}_{n}^{2}$•8^n-4+…+${C}_{n}^{n-2}$）．
再根据 ${C}_{n}^{0}$•8^n-2+${C}_{n}^{1}$•8^n-3+${C}_{n}^{2}$•8^n-4+…+${C}_{n}^{n-2}$为正整数，
可得3²ⁿ-8n-1能被64整除．

点评 本题考查二项式定理的应用：处理整除问题，利用函数的单调性证明不等式，体现了转化的数学思想，属于基础题．