已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561……求(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2的个位数字.
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![已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561……求(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2的个位数字.](/uploads/image/z/188725-13-5.jpg?t=%E5%B7%B2%E7%9F%A53%5E1%3D3%2C3%5E2%3D9%2C3%5E3%3D27%2C3%5E4%3D81%2C3%5E5%3D243%2C3%5E6%3D729%2C3%5E7%3D2187%2C3%5E8%3D6561%E2%80%A6%E2%80%A6%E6%B1%82%283-1%29%283%2B1%29%283%5E2%2B1%29%283%5E4%2B1%29%E2%80%A6%E2%80%A6%EF%BC%883%5E32%2B1%29%2B2%E7%9A%84%E4%B8%AA%E4%BD%8D%E6%95%B0%E5%AD%97.)
已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561……求(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2的个位数字.
已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561……求(3-1)(3+1)(3^2+1)
(3^4+1)……(3^32+1)+2的个位数字.
已知3^1=3,3^2=9,3^3=27,3^4=81,3^5=243,3^6=729,3^7=2187,3^8=6561……求(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2的个位数字.
解答﹙不断利用平方差公式﹚:
原式=﹙3²-1﹚﹙3²+1﹚……﹙3^32+1﹚+2
=﹙3^32-1﹚﹙3^32+1﹚+2
=3^64-1+2
=3^64+1
∵3^64的个位数=3^4个位数=1
∴原式的个位数字=2
原式=(3^2-1)(........
................
=3^64-1
64÷4=16∴3^64与3^4个位一样是1
∴原式的个位是1-1=0
(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2
=3^64-1+2
=3^64+1
=(3^4)^16+1
3^4个位数为1所以(3^4)^16个位数也为1
原式的个位数字为2
解:(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2=3^64-1+2=3^64+1
64能被4整除, 所以3^64个位数字是1, 1+1=2,
(3-1)(3+1)(3^2+1)(3^4+1)……(3^32+1)+2的个位数字是2.
此题用到平方差公式.