経済学で出る数学

ワークブックでじっくり攻める：問 4.17

1. ここでは，公式 \begin{align} \sum_{k=1}^{t}k = \dfrac{1}{2}t(t+1)\tag{#}\\\\ \sum_{k=1}^{t}k^2 = \dfrac{1}{6}t(t+1)(2t+1)\tag{♡}\\ \end{align} （『経出る』練習問題4.1，4.2）を使う． \[ \sum_{k=1}^{t}(k-1)k=\sum_{k=1}^{t}k^2- \sum_{k=1}^{t}k =\dfrac{1}{6}t(t+1)(2t+1)-\dfrac{1}{2}t(t+1) =\dfrac{1}{6}t(t+1)\left((2t+1)-3\right) =\dfrac{1}{3}t(t+1)(t-1)． \]


2. $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$ だから， \[ \sum_{k=1}^{t}\dfrac{1}{k(k+1)}=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots +\left(\dfrac{1}{t-1}-\dfrac{1}{t}\right) +\left(\dfrac{1}{t}-\dfrac{1}{t+1}\right)=1-\dfrac{1}{t+1}=\dfrac{t}{t+1}. \]