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ĐỀ THI Toán học
Số câu hỏi: 18
Thời gian làm bài: 32 phút
Mã đề: #1096
Lĩnh vực: Toán học
Nhóm: Toán 11 - Dãy số - cấp số cộng - cấp số nhân
Lệ phí: Miễn phí
Lượt thi: 3301

Ôn tập trắc nghiệm Dãy số Toán Lớp 11 Phần 3

Câu 1

 Cho dãy số \(\left(u_{n}\right) \text { vói }\left\{\begin{array}{l} u_{1}=1 \\ u_{n+1}=u_{n}+(-1)^{2n} \end{array}\right.\). Số hạng tổng quát \(u_n\) của dãy số là số hạng nào dưới đây?

A.
\(u_{n}=1+n\)
B.
\(u_{n}=1-n\)
C.
\(u_{n}=1+(-1)^{2 n}\)
D.
\(u_{n}=n\)
Câu 2

Cho dãy số \((u_n)\) un với \(\left\{\begin{array}{l} u_{1}=5 \\ u_{n+1}=u_{n}+n \end{array}\right.\).Số hạng tổng quát \(u_n\) của dãy số là số hạng nào dưới đây? 

A.
\(u_{n}=\frac{(n-1) n}{2}\)
B.
\(u_{n}=5+\frac{(n-1) n}{2}\)
C.
\(u_{n}=5+\frac{(n+1) n}{2}\)
D.
\(u_{n}=5+\frac{(n+1)(n+2)}{2}\)
Câu 3

Cho dãy số có các số hạng đầu là: \(\frac{1}{3} ; \frac{1}{3^{2}} ; \frac{1}{3^{3}} ; \frac{1}{3^{4}} ; \frac{1}{3^{5}};\cdots\).Số hạng tổng quát của dãy số này là?
 

A.
\(u_{n}=\frac{1}{3} \frac{1}{3^{n+1}}\)
B.
\(u_{n}=\frac{1}{3^{n+1}}\)
C.
\( u_{n}=\frac{1}{3^{n}}\)
D.
\(u_{n}=\frac{1}{3^{n-1}}\)
Câu 4

Cho dãy số có các số hạng đầu là: -2;0;2;4;6;....Số hạng tổng quát của dãy số này có dạng?

A.
\(u_{n}=-2 n\)
B.
\(u_{n}=(-2)+n\)
C.
\(u_{n}=(-2)(n+1)\)
D.
\(u_{n}=(-2)+2(n-1)\)
Câu 5

Cho dãy số có các số hạng đầu là:\(:-1 ; 1 ;-1 ; 1 ;-1 ; \ldots\)Số hạng tổng quát của dãy số này có dạng :

A.
\(u_{n}=1\)
B.
\(u_{n}=-1\)
C.
\(u_{n}=(-1)^{n}\)
D.
\(u_{n}=(-1)^{n+1}\)
Câu 6

Cho dãy số có các số hạng đầu là: 0,1; 0,01; 0,001; 0,0001;... . Số hạng tổng quát của dãy số này có dạng?

A.
\(u_{n}=\underbrace{0,00 \ldots 01}_{n \,chữ\, số\, 0}\)
B.
\(u_{n}=\underbrace{0,00 \ldots 01}_{n-1\,chữ\, số\, 0}\)
C.
\(\begin{aligned} &u_{n}=\frac{1}{10^{n-1}} \end{aligned}\)
D.
\(u_{n}=\frac{1}{10^{n+1}}\)
Câu 7

Cho dãy số có các số hạng đầu là:\(0 ; \frac{1}{2} ; \frac{2}{3} ; \frac{3}{4} ; \frac{4}{5},\cdots\) .Số hạng tổng quát của dãy số này là:

A.
\(u_{n}=\frac{n+1}{n}\)
B.
\(u_{n}=\frac{n}{n+1}\)
C.
\(u_{n}=\frac{n-1}{n}\)
D.
\(u_{n}=\frac{n^{2}-n}{n+1}\)
Câu 8

Cho dãy số có các số hạng đầu là:8,15, 22, 29,36,....Số hạng tổng quát của dãy số này là:

A.
\(u_{n}=7 n+7\)
B.
\(u_{n}=7 . n\)
C.
\(u_{n}=7 . n+1\)
D.
\(u_{n}\) Không viết được dưới dạng công thức
Câu 9

Cho dãy số có các số hạng đầu là: 5;10;15;20;25;... Số hạng tổng quát của dãy số này là

A.
\(u_{n}=5(n-1)\)
B.
\(u_{n}=5 n\)
C.
\(u_{n}=5+n\)
D.
\(u_{n}=5 . n+1\)
Câu 10

Cho dãy số \((u_n)\) với \(u_{n}=\frac{a n^{2}}{n+1}\)(a: hằng số).\(u_{n+1}\) là số hạng nào sau đây? 

A.
\(u_{n+1}=\frac{a \cdot(n+1)^{2}}{n+2}\)
B.
\(u_{n+1}=\frac{a \cdot(n+1)^{2}}{n+1}\)
C.
\(u_{n+1}=\frac{a \cdot n^{2}+1}{n+1}\)
D.
\(u_{n+1}=\frac{a n^{2}}{n+2}\)
Câu 11

Cho dãy số có 4 số hạng đầu là: -1,3,19,53. Hãy tìm  số hạng thứ 10 của dãy với quy luật vừa tìm.

A.
\(u_{10}=97\)
B.
\(u_{10}=71\)
C.
\(u_{10}=1414\)
D.
\(u_{10}=971\)
Câu 12

Cho dãy số \((u_n)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaIXaaabaGa % amyDamaaBaaaleaacaWGUbGaey4kaSIaaGymaaqabaGccqGH9aqpca % WG1bWaaSbaaSqaaiaad6gaaeqaaOGaey4kaSIaamOBamaaCaaaleqa % baGaaGOmaaaaaaGccaGL7baaaaa!4472! \left\{ \begin{array}{l} {u_1} = 1\\ {u_{n + 1}} = {u_n} + {n^2} \end{array} \right.\). Số hạng tổng quát của dãy số là số hạng nào dưới đây?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgUcaRiaaigdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!456B! {u_n} = 1 + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgUcaRiaaikdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4577! {u_n} = 1 + \frac{{n\left( {n - 1} \right)\left( {2n + 2} \right)}}{6}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgkHiTiaaigdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4581! {u_n} = 1 + \frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgkHiTiaaikdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4577! {u_n} = 1 + \frac{{n\left( {n + 1} \right)\left( {2n - 2} \right)}}{6}\)
Câu 13

Cho dãy số \((u_n)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaIXaaabaGa % amyDamaaBaaaleaacaWGUbGaey4kaSIaaGymaaqabaGccqGH9aqpca % WG1bWaaSbaaSqaaiaad6gaaeqaaOGaey4kaSYaaeWaaeaacqGHsisl % caaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaGaamOBaiabgU % caRiaaigdaaaaaaOGaay5Eaaaaaa!4940! \left\{ \begin{array}{l} {u_1} = 1\\ {u_{n + 1}} = {u_n} + {\left( { - 1} \right)^{2n + 1}} \end{array} \right.\). Số hạng tổng quát  của dãy số là số hạng nào dưới đây?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaikdacqGHsislcaWGUbaaaa!3BB9! {u_n} = 2 - n\)
B.
\(u_n\) không xác định 
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHsislcaWGUbaaaa!3BB8! {u_n} = 1 - n\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iabgkHiTiaad6gaaaa!3AFD! {u_n} = - n\) với mọi n 
Câu 14

Cho dãy số \((u_n)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpdaWcaaqaaiaa % igdaaeaacaaIYaaaaaqaaiaadwhadaWgaaWcbaGaamOBaiabgUcaRi % aaigdaaeqaaOGaeyypa0JaamyDamaaBaaaleaacaWGUbaabeaakiab % gkHiTiaaikdaaaGaay5Eaaaaaa!441E! \left\{ \begin{array}{l} {u_1} = \frac{1}{2}\\ {u_{n + 1}} = {u_n} - 2 \end{array} \right.\). Công thức số hạng tổng quát của dãy số này là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda % aaGaey4kaSIaaGOmamaabmaabaGaamOBaiabgkHiTiaaigdaaiaawI % cacaGLPaaaaaa!4065! {u_n} = \frac{1}{2} + 2\left( {n - 1} \right)\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda % aaGaeyOeI0IaaGOmamaabmaabaGaamOBaiabgkHiTiaaigdaaiaawI % cacaGLPaaaaaa!4070! {u_n} = \frac{1}{2} - 2\left( {n - 1} \right)\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda % aaGaeyOeI0IaaGOmaiaad6gaaaa!3D3F! {u_n} = \frac{1}{2} - 2n\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaaGymaaqaaiaaikda % aaGaey4kaSIaaGOmaiaad6gaaaa!3D34! {u_n} = \frac{1}{2} + 2n\)
Câu 15

Cho dãy số \((u_n)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcqGHsislcaaI % YaaabaGaamyDamaaBaaaleaacaWGUbGaey4kaSIaaGymaaqabaGccq % GH9aqpcqGHsislcaaIYaGaeyOeI0YaaSaaaeaacaaIXaaabaGaamyD % amaaBaaaleaacaWGUbaabeaaaaaaaOGaay5Eaaaaaa!45F9! \left\{ \begin{array}{l} {u_1} = - 2\\ {u_{n + 1}} = - 2 - \frac{1}{{{u_n}}} \end{array} \right.\). Công thức số hạng tổng quát của dãy số này là:

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iabgkHiTmaalaaabaGaamOBaiab % gkHiTiaaigdaaeaacaWGUbaaaaaa!3DA8! {u_n} = - \frac{{n - 1}}{n}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaamOBaiabgUcaRiaa % igdaaeaacaWGUbaaaaaa!3CB0! {u_n} = \frac{{n + 1}}{n}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iabgkHiTmaalaaabaGaamOBaiab % gUcaRiaaigdaaeaacaWGUbaaaaaa!3D9D! {u_n} = - \frac{{n + 1}}{n}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iabgkHiTmaalaaabaGaamOBaaqa % aiaad6gacqGHRaWkcaaIXaaaaaaa!3D9D! {u_n} = - \frac{n}{{n + 1}}\)
Câu 16

Cho dãy số \((u_n)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaGaba % abaeqabaGaamyDamaaBaaaleaacaaIXaaabeaakiabg2da9iaaikda % aeaacaWG1bWaaSbaaSqaaiaad6gacqGHRaWkcaaIXaaabeaakiabgk % HiTiaadwhadaWgaaWcbaGaamOBaaqabaGccqGH9aqpcaaIYaGaamOB % aiabgkHiTiaaigdaaaGaay5Eaaaabaaaaaa!45F6! \begin{array}{l} \left\{ \begin{array}{l} {u_1} = 2\\ {u_{n + 1}} - {u_n} = 2n - 1 \end{array} \right.\\ \end{array}\). Số hạng tổng quát \(u_n\) của dãy số là số hạng nào dưới đây?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaikdacqGHRaWkdaqadaqaaiaa % d6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaaaa!3FC8! {u_n} = 2 + {\left( {n - 1} \right)^2}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaikdacqGHRaWkcaWGUbWaaWba % aSqabeaacaaIYaaaaaaa!3C97! {u_n} = 2 + {n^2}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaikdacqGHRaWkdaqadaqaaiaa % d6gacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaaaa!3FBD! {u_n} = 2 + {\left( {n + 1} \right)^2}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaikdacqGHsisldaqadaqaaiaa % d6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaaaa!3FD3! {u_n} = 2 - {\left( {n - 1} \right)^2}\)
Câu 17

Cho dãy số \(u_n\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaIXaaabaGa % amyDamaaBaaaleaacaWGUbGaey4kaSIaaGymaaqabaGccqGH9aqpca % WG1bWaaSbaaSqaaiaad6gaaeqaaOGaey4kaSIaamOBamaaCaaaleqa % baGaaGOmaaaaaaGccaGL7baaaaa!4472! \left\{ \begin{array}{l} {u_1} = 1\\ {u_{n + 1}} = {u_n} + {n^2} \end{array} \right.\). Số hạng tổng quát \(u_n\) của dãy số là số hạng nào dưới đây?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgUcaRiaaigdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!456B! {u_n} = 1 + \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgUcaRiaaikdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4577! {u_n} = 1 + \frac{{n\left( {n - 1} \right)\left( {2n + 2} \right)}}{6}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHsislcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgkHiTiaaigdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4581! {u_n} = 1 + \frac{{n\left( {n - 1} \right)\left( {2n - 1} \right)}}{6}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaigdacqGHRaWkdaWcaaqaaiaa % d6gadaqadaqaaiaad6gacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaae % WaaeaacaaIYaGaamOBaiabgkHiTiaaikdaaiaawIcacaGLPaaaaeaa % caaI2aaaaaaa!4577! {u_n} = 1 + \frac{{n\left( {n + 1} \right)\left( {2n - 2} \right)}}{6}\)
Câu 18

Cho dãy số \(u_n\) với  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaI1aaabaGa % amyDamaaBaaaleaacaWGUbGaey4kaSIaaGymaaqabaGccqGH9aqpca % WG1bWaaSbaaSqaaiaad6gaaeqaaOGaey4kaSIaamOBaaaacaGL7baa % aaa!4383! \left\{ \begin{array}{l} {u_1} = 5\\ {u_{n + 1}} = {u_n} + n \end{array} \right.\) Số hạng tổng quát \(u_n\) của dãy số là số hạng nào dưới đây?

A.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9maalaaabaGaaiikaiaad6gacqGH % sislcaaIXaGaaiykaiaad6gaaeaacaaIYaaaaaaa!3ED0! {u_n} = \frac{{(n - 1)n}}{2}\)
B.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaiwdacqGHRaWkdaWcaaqaaiaa % cIcacaWGUbGaeyOeI0IaaGymaiaacMcacaWGUbaabaGaaGOmaaaaaa % a!4071! {u_n} = 5 + \frac{{(n - 1)n}}{2}\)
C.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaiwdacqGHRaWkdaWcaaqaaiaa % cIcacaWGUbGaey4kaSIaaGymaiaacMcacaWGUbaabaGaaGOmaaaaaa % a!4066! {u_n} = 5 + \frac{{(n + 1)n}}{2}\)
D.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaWGUbaabeaakiabg2da9iaaiwdacqGHRaWkdaWcaaqaaiaa % cIcacaWGUbGaey4kaSIaaGymaiaacMcacaGGOaGaamOBaiabgUcaRi % aaikdacaGGPaaabaGaaGOmaaaaaaa!435D! {u_n} = 5 + \frac{{(n + 1)(n + 2)}}{2}\)