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數列總結

Authored by iokhong che

Mathematics

9th - 12th Grade

Used 43+ times

數列總結
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10 questions

Show all answers

1.

MATCH QUESTION

2 mins • 5 pts

等差數列公式配對

Sn=(a1+an)×n2S_n=\frac{\left(a_1+a_n\right)\times n}{2}

性質

m+n=p+q⟺am+an=ap+aqm+n=p+q\Longleftrightarrow a_m+a_n=a_p+a_q

求和公式1

an=a1+(n−1)da_n=a_1+\left(n-1\right)d

通項公式

A=a+b2A=\frac{a+b}{2}

求和公式2

Sn=na1+n(n−1)×d2S_n=na_1+\frac{n\left(n-1\right)\times d}{2}

等差中項

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

已知數列 {an}\left\{a_n\right\} 的前 nn 項和 Sn=n2+3nS_n=n^2+3n ,則通項公式 an=a_n=

an=2n−2a_n=2n-2

an=2n+2a_n=2n+2

an=n2−2na_n=n^2-2n

an=n2+2na_n=n^2+2n

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

已知 a=13+2a=\frac{1}{\sqrt[]{3}+\sqrt[]{2}} , b=13−2b=\frac{1}{\sqrt[]{3}-\sqrt[]{2}} ,則 a,ba,b 的等差中項=

3\sqrt[]{3}

2\sqrt[]{2}

33\frac{\sqrt[]{3}}{3}

22\frac{\sqrt[]{2}}{2}

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

已知等差數列 {an}\left\{a_n\right\} ,且 a1+a2+a3=18a_1+a_2+a_3=18 ,則 a2=a_2=

5

7

6

8

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

在等差數列 {an}\left\{a_n\right\} 中,已知它的前14項之和為28,則 a5+a10=a_5+a_{10}=

14

28

8

4

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

已知等差數列 {an}\left\{a_n\right\} 中, a1=9,a3+a8=0a_1=9,a_3+a_8=0 ,則通項公式 an=a_n=

11−2n11-2n

7+2n7+2n

7−2n7-2n

11+2n11+2n

7.

MATCH QUESTION

2 mins • 1 pt

等比數列相關公式

Sn=a1−anr1−r (r≠1)S_n=\frac{a_1-a_nr}{1-r}\ \left(r\ne1\right)

性質

an=a1rn−1a_n=a_1r^{n-1}

求和公式2

Sn=a11−r (∣r∣<1)S_n=\frac{a_1}{1-r}\ \left(\left|r\right|<1\right)

無窮遞縮等比數列求和

m+n=l+p⟺aman=alapm+n=l+p\Longleftrightarrow a_ma_n=a_la_p

求和公式1

Sn=a1(1−rn)1−r  (r≠1)S_n=\frac{a_1\left(1-r^n\right)}{1-r}\ \ \left(r\ne1\right)

通項公式

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