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標題:
prove by factor theorem that 9^n-1 is divisible by 8
發問:
prove by factor theorem that (9^n)-1 is divisible by 8 for all positive integers n. 更新: 沒錯這是個可行的方法 可是你的答案不合要求
最佳解答:
Let f(x)=9^n-1 By applying the formula a^n-b^n=(a-b)[a^(n-1)+a^(n-2)b+......ab^(n-2)+b^(n-1)] f(x)=9^n-1=8[9^(n-1)+9^(n-2)+......9+1], which is divisible by 8 no matter what value of n is. [我唔知咁樣算唔算用左factor theorem，希望幫到你啦~~]
prove by factor theorem that 9^n-1 is divisible by 8
發問:
prove by factor theorem that (9^n)-1 is divisible by 8 for all positive integers n. 更新: 沒錯這是個可行的方法 可是你的答案不合要求
最佳解答:
Let f(x)=9^n-1 By applying the formula a^n-b^n=(a-b)[a^(n-1)+a^(n-2)b+......ab^(n-2)+b^(n-1)] f(x)=9^n-1=8[9^(n-1)+9^(n-2)+......9+1], which is divisible by 8 no matter what value of n is. [我唔知咁樣算唔算用左factor theorem，希望幫到你啦~~]
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