(a) 通过考虑 $(r+1)^2-r^2$，用差分法证明 $\displaystyle\sum_{r=1}^{n}r=\frac12n(n+1)$。

(b) 已知 $\displaystyle\sum_{r=1}^{n}(r+a)=n$，求 $a$，以 $n$ 表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

把相邻平方差展开为2r+1，对r求和后，左边首尾相消，右边包含所求自然数和。

Since $(r+1)^2-r^2=2r+1$, summing from $r=1$ to $r=n$ gives

The left-hand side telescopes to $(n+1)^2-1=n^2+2n$. Thus

so

The expansion illustrates the cancellation for sufficiently large $n$; the identity also holds directly for smaller positive integers.

常数a加了n次，所以先用上一问的自然数和，再除以正整数n。

Using (a),

Since $n>0$, division by $n$ gives $\frac12(n+1)+a=1$. Hence