1.(1) n(n+1)(n+2)(n+3)+1=[n*(n+3)+1]^2
(2) 证：左边=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2=右边
2.(1) 4294967296
(2) 原式=(2-1)(2+1)(2^2+1)(2^3+1)……(2^32+1)
=2^64-1
个位数字2,4,8,6循环一次,所以2^64个位是6
故个位是5
1.(1) n(n+1)(n+2)(n+3)+1=[n*(n+3)+1]^2
(2) 证：左边=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2=右边
2.(1) 4294967296
(2) 原式=(2-1)(2+1)(2^2+1)(2^3+1)……(2^32+1)
=2^64-1
个位数字2,4,8,6循环一次,所以2^64个位是6
故个位是5