用数学归纳法证明:“(n+1)*(n+2)*…*(n+n)=2^n*1*3*…*(2n-1)”.从“k到k+1”左端需增乘的代数式为( )(A)2k+1 (B)2(2k+1)(C)(2k+1)/(k+1) (D)(2k+3)/(k+1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 08:51:41
![用数学归纳法证明:“(n+1)*(n+2)*…*(n+n)=2^n*1*3*…*(2n-1)”.从“k到k+1”左端需增乘的代数式为( )(A)2k+1 (B)2(2k+1)(C)(2k+1)/(k+1) (D)(2k+3)/(k+1)](/uploads/image/z/7302975-15-5.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%E2%80%9C%EF%BC%88n%2B1%EF%BC%89%2A%EF%BC%88n%2B2%EF%BC%89%2A%E2%80%A6%2A%EF%BC%88n%2Bn%EF%BC%89%3D2%5En%2A1%2A3%2A%E2%80%A6%2A%EF%BC%882n-1%EF%BC%89%E2%80%9D.%E4%BB%8E%E2%80%9Ck%E5%88%B0k%2B1%E2%80%9D%E5%B7%A6%E7%AB%AF%E9%9C%80%E5%A2%9E%E4%B9%98%E7%9A%84%E4%BB%A3%E6%95%B0%E5%BC%8F%E4%B8%BA%EF%BC%88+%EF%BC%89%EF%BC%88A%EF%BC%892k%2B1+%EF%BC%88B%EF%BC%892%EF%BC%882k%2B1%EF%BC%89%EF%BC%88C%EF%BC%89%EF%BC%882k%2B1%EF%BC%89%2F%EF%BC%88k%2B1%EF%BC%89+%EF%BC%88D%EF%BC%89%EF%BC%882k%2B3%EF%BC%89%2F%EF%BC%88k%2B1%EF%BC%89)
用数学归纳法证明:“(n+1)*(n+2)*…*(n+n)=2^n*1*3*…*(2n-1)”.从“k到k+1”左端需增乘的代数式为( )(A)2k+1 (B)2(2k+1)(C)(2k+1)/(k+1) (D)(2k+3)/(k+1)
用数学归纳法证明:“(n+1)*(n+2)*…*(n+n)=2^n*1*3*…*(2n-1)”.从“k到k+1”左端需增乘的代数式为( )
(A)2k+1 (B)2(2k+1)
(C)(2k+1)/(k+1) (D)(2k+3)/(k+1)
用数学归纳法证明:“(n+1)*(n+2)*…*(n+n)=2^n*1*3*…*(2n-1)”.从“k到k+1”左端需增乘的代数式为( )(A)2k+1 (B)2(2k+1)(C)(2k+1)/(k+1) (D)(2k+3)/(k+1)
B(1)当n=1时,(1+1)=2=2^1*1等式成立;(2)假设n=k成立,即(k+1)(k+2)~(k+k)=2^k*1*3*~*(2k-1)成立,则n=k+1时,左式=(k+1+1)(k+1+2)~(k+1+k-2)(k+1+k-1)(k+1+k)(k+1+k+1)=(k+2)(k+3)~(2k-1)(k+k)(2k+1)2(k+1)=(k+1)(k+2)~(k+k)2(2k+1)=2^k*1*3*~*(2k-1)*2*(2k+1)=2^(k+1)*1*3*~*(2k+1)=右式,等式成立
答案是B。<(k+1)+1><(k+1)+2>~~<(k+1)+(K+1)>=(k+1)(k+2)~~(k+k).(k+k+1)(k+k+2)/(k+1)=~~~~~~(2k+1)(2k+2)/(k+1)=~~~~~~~(2k+1).2
少了k+1项,多了(k+1+k)(k+1+k+1),故增乘的代数式为(k+1+k)(k+1+k+1)/(k+1)=2(2k+1)
选B