GWD 21-2

lilyblue_zl

The sum of the first k positive integers is equal to k(k+1)/2. What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. m(m+1)/2-(n+1)(n+2)/2

B. m(m+1)/2-n(n+1)/2

C. m(m+1)/2-(n-1)n/2

D. (m-1)m/2-(n+1)(n+2)/2

E. (m-1)m/2-n(n+1)/2

The answer is C

Please help to explain it

诡异儿

怎会是加n

应该是减吧

原式0<n<m 求n到m, 过程重复计算了n, 因此要减掉

变成: [n(n+1)/2]-n 通分后n变成2n/2

解开=(n^2+n-2n)/2 = (n^2-n)/2

提出n=n(n-1)/2

hugewaters

多加个N就是了,,

qianrener

我怎么愣是没看懂呢? 哪位可以帮解释下吗?

lilyblue_zl

I see, thanks a ton

罗杰Liu

m(m+1)/2-(n-1)n/2＝[m(m+1)/2-n(n+1)/2]+n

lz一定是忘了应该加上n，不过我第一次做的话也很有可能会忘的