用数学归纳法证明3^4n+1+5^2n+1],n为整数能被8整除时,当N=k+1时,对于3^{4[k+1]+1}+5^{2[k+1]

Question

用数学归纳法证明3^【4n+1】+5^【2n+1],【n为整数】能被8整除时,当N=k+1时,对于3^{4[k+1]+1}+5^{2[k+1]

Answer 1

3^{4[k+1]+1}+5^{2[k+1]+1}
=3^{(4k+1)+4}+5^{(2k+1)+2}
={3^(4k+1)*3^4}+{5^(2k+1)*5^2}
=81*3^(4k+1)+25*5^(2k+1)
=25*{3^(4k+1)+25*5^(2k+1)}+56*3^(4k+1)
3^(4k+1)+25*5^(2k+1)能被8整除,所以25*3^(4k+1)+25*5^(2k+1)能被8整除.
56*3^(4k+1)能被8整除
所以25*{3^(4k+1)+25*5^(2k+1)}+56*3^(4k+1)能被8整除.
3^{4[k+1]+1}+5^{2[k+1]+1}能被8整除