设S=1+2+3+...+n,则S=n+(n-1)+(n-2)+...+1.由上得2S =(n+1)+(n+1)+(n+
5个回答
设S=1+3+5+.+(2n-1)
则S=(2n-1)+(2n-3)+.+3+2+1
两式顺序相加
2s=2n+2n+.+2n (共n个)
2n^2
∴1+3+5+...+(2n-1)=n的平方
相关问题
设S(n)=1/n+1/n+1+1/n+2+1/n+3+……1/n^2 当n=2时,S(2)
设S=1·1!+2·2!+3·3!+…+n·n!则S=?
若 S n = 1 1•2 + 1 2•3 + 1 3•4 …+ 1 n•(n+1) (n∈ N * ) ,则S 10
设数列{an}的前n项和为Sn,且满足S1=2,S(n+1)=3S(n+2)(n=1,2,3) 设bn=2,S(n+1)
设数列{a n }的前n项和为S n ,已知a 1 =1,S n+1 =2S n +n+1(n∈N*),
数列{An}满足a(1)=1,a(2)=6,S(n)=3S(n-1)-2S(n-2)+2^n(n>=3)
n=1 s=1 n=2 s=3 n=3 s=6 n=4 s=10
设数列{a(n)}的前n项和为S(n)·已知S(1)=1,S(n+1)/S(n)=n+2/n·且a(1),a(2),a(
设数列{a n }的前n项和为S n ,已知a 1 =2,a 2 =8,S n+1 +4S n-1 =5S n (n≥2
已知数列{a n }的前n项和为S n ,若S 1 =1,S 2 =2,且S n+1 -3S n +2S n-1 =0(