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 （\(\sum\limits_{i&#x3D;0}^n\text{f}(i)\) 表示从 \(0\) 到 \(n\) 对 \(\text{f}(i)\) 求和）
 （约定 \(\text{C}^0_n&#x3D;1\)）
 
 引理：n 次方差公式 \[(m+1)^{k+1}-m^{k+1}&#x3D;\sum_{i&#x3D;0"><meta property="og:type" content="article"><meta property="og:title" content="自然数幂求和公式"><meta property="og:url" content="https://schtonn.github.io/blog/posts/powersum/index.html"><meta property="og:site_name" content="schtonn"><meta property="og:description" content="（\(\text{C}^m_n\) 代表从 \(n\) 个不同物品里面选 \(m\) 个，有多少种选法）
 （\(\sum\limits_{i&#x3D;0}^n\text{f}(i)\) 表示从 \(0\) 到 \(n\) 对 \(\text{f}(i)\) 求和）
 （约定 \(\text{C}^0_n&#x3D;1\)）
 
 引理：n 次方差公式 \[(m+1)^{k+1}-m^{k+1}&#x3D;\sum_{i&#x3D;0"><meta property="og:locale" content="en_US"><meta property="article:published_time" content="2021-12-12T04:24:01.513Z"><meta property="article:modified_time" content="2022-10-19T15:10:22.174Z"><meta property="article:author" content="Alex"><meta property="article:tag" content="math"><meta name="twitter:card" content="summary"><link rel="canonical" href="https://schtonn.github.io/blog/posts/powersum/"><script id="page-configurations">CONFIG.page={sidebar:"",isHome:!1,isPost:!0,lang:"en"}</script><title>自然数幂求和公式 | schtonn</title><noscript><style>.sidebar-inner,.use-motion .brand,.use-motion .collection-header,.use-motion .comments,.use-motion .menu-item,.use-motion .pagination,.use-motion .post-block,.use-motion .post-body,.use-motion .post-header{opacity:initial}.use-motion .site-subtitle,.use-motion .site-title{opacity:initial;top:initial}.use-motion .logo-line-before i{left:initial}.use-motion .logo-line-after 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class="post-meta-item-text">Edited on</span> <time title="Modified: 2022-Oct-19 23:10:22" itemprop="dateModified" datetime="2022-10-19T23:10:22+08:00">2022-Oct-19</time></span><span class="post-meta-item"><span class="post-meta-item-icon"><i class="fa fa-comment-o"></i></span> <span class="post-meta-item-text">Valine:</span><a title="valine" href="/blog/posts/powersum/#valine-comments" itemprop="discussionUrl"><span class="post-comments-count valine-comment-count" data-xid="/blog/posts/powersum/" itemprop="commentCount"></span></a></span></div></header><div class="post-body" itemprop="articleBody"><p>（<span class="math inline">\(\text{C}^m_n\)</span> 代表从 <span class="math inline">\(n\)</span> 个不同物品里面选 <span class="math inline">\(m\)</span> 个，有多少种选法）</p><p>（<span class="math inline">\(\sum\limits_{i=0}^n\text{f}(i)\)</span> 表示从 <span class="math inline">\(0\)</span> 到 <span class="math inline">\(n\)</span> 对 <span class="math inline">\(\text{f}(i)\)</span> 求和）</p><p>（约定 <span class="math inline">\(\text{C}^0_n=1\)</span>）</p><blockquote><p>引理：n 次方差公式 <span class="math display">\[(m+1)^{k+1}-m^{k+1}=\sum_{i=0}^{k}\text{C}^i_{k+1}m^i\]</span></p></blockquote><p>证：</p><p><span class="math display">\[\sum^n_{m=1}(m+1)^{k+1}-m^{k+1}=(n+1)^{k+1}-1^{k+1}\]</span> <span class="math display">\[=\sum_{i=0}^k\text{C}^i_{k+1}\left(\sum^n_{m=1}m^i\right)\]</span> <span class="math display">\[=\text{C}^k_{k+1}\left(\sum^n_{m=1}m^k\right)+\sum^{k-1}_{i=0}\text{C}^i_{k+1}\left(\sum^n_{m=1}m^i\right)\]</span></p><p>因此有 <span class="math display">\[\sum^n_{m=1}m^k=\dfrac{1}{\text{C}^k_{k+1}}\left[(n+1)^{k+1}-1-\sum^{k-1}_{i=0}\text{C}^i_{k+1}\left(\sum^n_{m=1}m^i\right)\right]\]</span></p></div><div><ul class="post-copyright"><li class="post-copyright-author"> <strong>Post author:</strong> Alex</li><li class="post-copyright-link"> <strong>Post link:</strong> <a href="https://schtonn.github.io/blog/posts/powersum/" 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