一道简单的GRE数学题题干:1<n<5,n is an integer,比较大小A:the sum of the first n odd integer that are greater then zeroB:n^2-1(n的平方减一)答案是A大于B

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一道简单的GRE数学题题干:1<n<5,n is an integer,比较大小A:the sum of the first n odd integer that are greater then zeroB:n^2-1(n的平方减一)答案是A大于B

一道简单的GRE数学题题干:1<n<5,n is an integer,比较大小A:the sum of the first n odd integer that are greater then zeroB:n^2-1(n的平方减一)答案是A大于B
一道简单的GRE数学题
题干:1<n<5,n is an integer,比较大小
A:the sum of the first n odd integer that are greater then zero
B:n^2-1(n的平方减一)
答案是A大于B

一道简单的GRE数学题题干:1<n<5,n is an integer,比较大小A:the sum of the first n odd integer that are greater then zeroB:n^2-1(n的平方减一)答案是A大于B
n可以是2,3,4
A:前n个比零大的奇数的和
n=2,结果1+3=4.B:3
n=3,结果1+3+5=9 B:8
……
也即:A:(1+2n-1)*n/2=n^2
B:n^2-1
所以显然A比B大~