《2025年学易优同步学案导学高中数学选择性必修第二册通用版》

【跟踪训练4】 设$S_{n}$为等差数列$\{ a_{n}\}$的前$n$项和，已知$S_{3}=a_{7}$，$a_{8}-2a_{3}=3$.
(1)求$a_{n}$；
(2)设$b_{n}=\frac{1}{S_{n}}$，求数列$\{ b_{n}\}$的前$n$项和$T_{n}$.

答案： 解：
(1)设数列$\{a_{n}\}$的公差为d，由题意得$\begin{cases} 3a_{1}+3d=a_{1}+6d,\a_{1}+7d)-2(a_{1}+2d)=3,\end{cases}$解得$a_{1}=3，$d=2，$\therefore a_{n}=a_{1}+(n-1)d=2n+1。$
(2)由
(1)得$S_{n}=na_{1}+\frac{n(n-1)}{2}d=n(n+2)，$$\therefore b_{n}=\frac{1}{n(n+2)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})。$$\therefore T_{n}=b_{1}+b_{2}+·s+b_{n-1}+b_{n}=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+·s+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})]=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2})=\frac{3}{4}-\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n+2})。$

1.求值：$1 - 3 + 5 - 7 + 9 - 11 + ·s + 2025 - 2027$等于 (B)

A.$-2026$
B.$-1014$
C.$-507$
D.$1014$

答案： 1.B $1-3+5-7+9-11+·s+2025-2027=(1$$-3)+(5-7)+(9-11)+·s+(2025-2027)$$=-2 × 507=-1014$。

2.数列$1\frac{1}{2},3\frac{1}{4},5\frac{1}{8},7\frac{1}{16}·s$的前$n$项和$S_{n}$为 (C)

A.$n^{2}+1-\frac{1}{2^{n - 1}}$
B.$n^{2}+2-\frac{1}{2^{n}}$
C.$n^{2}+1-\frac{1}{2^{n}}$
D.$n^{2}+2-\frac{1}{2^{n - 1}}$

答案： 2.C 数列$1\frac{1}{2}$，$3\frac{1}{4}$，$5\frac{1}{8}$，$7\frac{1}{16}·s$的通项公式为$a_{n}$$=2n-1+(\frac{1}{2})^{n}$，所以$S_{n}=(1+\frac{1}{2})+(3+\frac{1}{2^{2}})+(5+\frac{1}{2^{3}})+$$(7+\frac{1}{2^{4}})+·s+(2n-1+\frac{1}{2^{n}})=[1+3+5+·s+$$(2n-1)]+(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+·s+\frac{1}{2^{n}})$$=\frac{n(1+2n-1)}{2}+\frac{\frac{1}{2}(1-\frac{1}{2^{n}})}{1-\frac{1}{2}}$$=n^{2}+1-\frac{1}{2^{n}}$。

3.已知数列$\{ a_{n}\}:\frac{1}{2},\frac{1}{3}+\frac{2}{3},\frac{1}{4}+\frac{2}{4}+\frac{3}{4},\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5},·s$，那么数列$\{ b_{n}\} = \left\{ \frac{1}{a_{n}a_{n + 1}} \right\}$前$n$项的和为 (A)

A.$4\left(1 - \frac{1}{n + 1}\right)$
B.$4\left(\frac{1}{2} - \frac{1}{n + 1}\right)$
C.$1 - \frac{1}{n + 1}$
D.$\frac{1}{2} - \frac{1}{n + 1}$

答案： 3.A $\because a_{n}=\frac{1+2+3+·s+n}{n+1}=\frac{\frac{n(n+1)}{2}}{n+1}=\frac{n}{2}$，$\therefore b_{n}=\frac{1}{a_{n}a_{n+1}}=\frac{4}{n(n+1)}=4(\frac{1}{n}-\frac{1}{n+1})$。$\therefore S_{n}=4(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+·s+\frac{1}{n}$$-\frac{1}{n+1})=4(1-\frac{1}{n+1})$。

4.$\sin^{2}1^{\circ}+\sin^{2}2^{\circ}+\sin^{2}3^{\circ}+·s+\sin^{2}88^{\circ}+\sin^{2}89^{\circ}=\underline{\underline{$44.5$}}$.

答案： 4.44.5 设$S=\sin^{2}1^{\circ}+\sin^{2}2^{\circ}+\sin^{2}3^{\circ}+·s+$$\sin^{2}88^{\circ}+\sin^{2}89^{\circ}$，①将①式右边反序得，$S=\sin^{2}89^{\circ}+\sin^{2}88^{\circ}+·s+\sin^{2}3^{\circ}+\sin^{2}2^{\circ}+$$\sin^{2}1^{\circ}$，②①+②得，$2S=(\sin^{2}1^{\circ}+\sin^{2}89^{\circ})+(\sin^{2}2^{\circ}+\sin^{2}88^{\circ})+·s$$+(\sin^{2}89^{\circ}+\sin^{2}1^{\circ})=(\sin^{2}1^{\circ}+\cos^{2}1^{\circ})+(\sin^{2}2^{\circ}$$+\cos^{2}2^{\circ})+·s+(\sin^{2}89^{\circ}+\cos^{2}89^{\circ})=89$，$\therefore S=44.5$。