余數(shù)為0 整除31
n=1時(shí)
1+2^1+2^2+2^3+4^2=1+2+4+8+16=31
成立
設(shè)k成立
1+...+2^(5k-1)=31p
k+1時(shí)
1+..+2^(5k-1)+2^5k+2^(5k+1)+2^(5k+2)+2^(5k+3)+2^(5(k+1)-1)
=31p+2^5k(1+2^1+2^2+2^3+2^4)
=31(p+2^5k)
k+1時(shí)成立