用数学归纳法证明:当整数n≥0时,(x+2)^(2n+2)-(x+1)^(n+1)能被x^2+3x+3整除?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 21:33:17
![用数学归纳法证明:当整数n≥0时,(x+2)^(2n+2)-(x+1)^(n+1)能被x^2+3x+3整除?](/uploads/image/z/6943001-41-1.jpg?t=%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%EF%BC%9A%E5%BD%93%E6%95%B4%E6%95%B0n%E2%89%A50%E6%97%B6%2C%EF%BC%88x%2B2%29%5E%282n%2B2%29-%28x%2B1%29%5E%28n%2B1%29%E8%83%BD%E8%A2%ABx%5E2%2B3x%2B3%E6%95%B4%E9%99%A4%3F)
用数学归纳法证明:当整数n≥0时,(x+2)^(2n+2)-(x+1)^(n+1)能被x^2+3x+3整除?
用数学归纳法证明:当整数n≥0时,(x+2)^(2n+2)-(x+1)^(n+1)能被x^2+3x+3整除?
用数学归纳法证明:当整数n≥0时,(x+2)^(2n+2)-(x+1)^(n+1)能被x^2+3x+3整除?
1)当整数n=0时,(x+2)^2-(x+1)=(x^2+4x+4)-x-1=x^2+3x+3能被x^2+3x+3整除
2)假设当整数n=K时,命题成立,即:(x+2)^(2K+2)-(x+1)^(K+1)能被x^2+3x+3整除
那么当整数n=K+1时:(x+2)^(2K+2+2)-(x+1)^(K+1+1)
=[(x+2)^(2K+2)]*(x+2)-[(x+1)^(k+1)]*(x+1)
=[(x+2)^(2K+2)-(x+1)^(K+1)]*(x+2)²-[(x+1)^(k+1)]*(x+1)+(x+1)^(K+1)*(x+2)²
=[(x+2)^(2K+2)-(x+1)^(K+1)]*(x+2)²-[(x+1)^(k+1)]*[(x+2)²-(x+1)]
=[(x+2)^(2K+2)-(x+1)^(K+1)]*(x+2)²-[(x+1)^(k+1)]*[(x²+3x+3)]
因为[(x+2)^(2K+2)-(x+1)^(K+1)]*(x+2)²能被x^2+3x+3整除
[(x+1)^(k+1)]*[(x²+3x+3)]也能被x^2+3x+3整除
所以(x+2)^(2K+2+2)-(x+1)^(K+1+1)能被x^2+3x+3整除
说明当整数n=K+1时命题成立,
由(1)(2)可知,命题对于n≥0整数都成立.