证明题练习2 数列

If $p,q$p,q are distinct natural numbers, and in an arithmetic sequence $\{a_n\}$, $S_p=S_q$, where $S_k$ denotes the sum of the first $k$ terms of the sequence $\{a_n\}$. Prove that $S_{p+q}=0$ .

证明（Proof）

For an arithmetic sequence $\{a_n\}${an​}, let the first term be $a_1$a1​ and the common difference be $d$d. The formula for the sum of the first $n$n terms is:
$S_n=n{a}_1+\frac{n(n-1)}{2}d$Sn​=na1​+2n(n−1)​d

Step 1: Use the condition $S_p=S_q$Sp​=Sq​

By the sum formula, substitute $n=p$n=p and $n=q$n=q:
$p{a}_1+\frac{p(p-1)}{2}d=q{a}_1+\frac{q(q-1)}{2}d$pa1​+2p(p−1)​d=qa1​+2q(q−1)​d

Rearrange terms to group $a_1$a1​ and $d$d:
$(p-q)a_1+\frac{p(p-1)-q(q-1)}{2}d=0$(p−q)a1​+2p(p−1)−q(q−1)​d=0

Step 2: Simplify the coefficient of $d$d

Factor the quadratic expression $p(p-1)-q(q-1)$p(p−1)−q(q−1):
$p(p-1)-q(q-1)=p^2-p-q^2+q=(p^2-q^2)-(p-q)$p(p−1)−q(q−1)=p2−p−q2+q=(p2−q2)−(p−q)
Using the difference of squares $p^2-q^2=(p-q)(p+q)$p2−q2=(p−q)(p+q), we get:
$(p^2-q^2)-(p-q)=(p-q)(p+q)-(p-q)=(p-q)(p+q-1)$(p2−q2)−(p−q)=(p−q)(p+q)−(p−q)=(p−q)(p+q−1)

Step 3: Substitute back and simplify

Substitute the factored form into the equation from Step 1:
$(p-q)a_1+\frac{(p-q)(p+q-1)}{2}d=0$(p−q)a1​+2(p−q)(p+q−1)​d=0

Since $p\mathrm{\ne }q$p=q (given that $p,q$p,q are distinct), $p-q\mathrm{\ne }0$p−q=0. Divide both sides by $p-q$p−q:
$a_1+\frac{(p+q-1)}{2}d=0$a1​+2(p+q−1)​d=0

Step 4: Compute $S_{p+q}$Sp+q​

Use the sum formula for $n=p+q$n=p+q:
$S_{p+q}=(p+q)a_1+\frac{(p+q)(p+q-1)}{2}d$Sp+q​=(p+q)a1​+2(p+q)(p+q−1)​d

Factor out $(p+q)$(p+q):
$S_{p+q}=(p+q)[a_1+\frac{(p+q-1)}{2}d]$Sp+q​=(p+q)[a1​+2(p+q−1)​d]

From Step 3, we know $a_1+\frac{(p+q-1)}{2}d=0$a1​+2(p+q−1)​d=0. Substitute this into the expression for $S_{p+q}$Sp+q​:
$S_{p+q}=(p+q)\cdot 0=0$Sp+q​=(p+q)⋅0=0

Thus, $S_{p+q}=0$Sp+q​=0 is proven.