用数学归纳法证明:(n+1)+(n+2)+…+(n+n)=n(3n+1)2(n∈N*)

问题描述:

用数学归纳法证明:(n+1)+(n+2)+…+(n+n)=

n(3n+1)
2
(n∈N*

证明:①n=1时,左边=2,右边=2,等式成立;
②假设n=k时,结论成立,即:(k+1)+(k+2)+…+(k+k)=

k(3k+1)
2

则n=k+1时,等式左边=(k+2)+(k+3)+…+(k+k+1)+(k+1+k+1)=
k(3k+1)
2
+3k+2=
(k+1)(3k+4)
2

故n=k+1时,等式成立
由①②可知:(n+1)+(n+2)+…+(n+n)=
n(3n+1)
2
(n∈N*)成立