(a)$\displaystyle\sum_{i=m}^na_i+\sum_{i=n+1}^pa_i=\sum_{i=m}^pa_i$.其中$m,n,q\in\mathbb{Z},$$m\leq n< p$.
证明：可见$n+1\leq p$.当$p=n+1$时，易得$\displaystyle\sum_{i=m}^na_i+a_{n+1}=\sum_{m}^{n+1}a_i$成立.假设当$p=k(n+1\leq k)$时，命题也成立，即$\displaystyle \sum_{i=m}^na_i+\sum_{i=n+1}^ka_i=\sum_{i=m}^ka_i$.则当$p=k+1$时，$$\sum_{m}^na_i+\sum_{i=n+1}^{k+1}a_i=\sum_{m}^na_i+(\sum_{i=n+1}^ka_i+a_{k+1})=(\sum_{m}^na_i+\sum_{i=n+1}^ka_i)+a_{k+1}=\sum_{m}^ka_i+a_{k+1}=\sum_{m}^{k+1}a_i$$根据数学归纳法，可见命题对于一切整数$p\succ n$成立.
引理7.1.4(c)
设$m\leq n$是整数，并设$a_i,b_i$是对应于每个整数$m\leq i\leq n$的实数.那么我们有
$$\sum_{i=m}^n(a_i+b_i)=\sum_{i=m}^na_i+\sum_{i=m}^nb_i$$
证明：当$m=n$时，显然$a_m+b_m=a_m+b_m$.设当$n=k(m\leq k)$时，命题成立，即
$$\sum_{i=m}^k(a_i+b_i)=\sum_{i=m}^ka_i+\sum_{i=m}^kb_i$$
则当$n=k+1$时，
$$\sum_{i=m}^{k+1}(a_i+b_i)=\sum_{i=m}^{k}(a_i+b_i)+(a_{k+1}+b_{k+1})=(\sum_{i=m}^ka_i+\sum_{i=m}^kb_i)+(a_{k+1}+b_{k+1})=(\sum_{i=m}^ka_i+a_{k+1})+(\sum_{i=m}^kb_i+b_{k+1})=\sum_{i=m}^{k+1}a_i+\sum_{i=m}^{k+1}b_i$$
根据数学归纳法，命题对于一切整数$n\geq m$成立.