Question

用数学归纳法证明：(n+1)(n+2)(n+3)+.+(n+n)=(2^n)*1*3*.(2n-1),从K到K+1,左端

Answer 1

证明：

n=1时,n+1=2

(2^1)*1=2,等式成立.

假设当n=k(k为自然数,且k>=1)时等式成立.

即

(k+1)(k+2)...(k+k)=(2^k)*1*3*...*(2k-1)

则当n=k+1时,

(k+1+1)(k+1+2)...(k+1+k-1)(k+1+k)(k+1+k+1)

=(k+2)(k+3)...(k+k)(2k+1)(2k+2)

=(k+1)(k+2)...(k+k)(2k+1)(2k+2)/(k+1)

=(k+1)(k+2)...(k+k)(2k+1)2

=(2^k)*1*3*...*(2k-1)*(2k+1)*2

=[2^(k+1)]*1*3*...*[2(k+1)-1]

等式也成立.

综上,(n+1)(n+2)(n+3)+.+(n+n)=(2^n)*1*3*.(2n-1)

等式成立.