A = 1( 0 + 1   ) + 2 ( 1 + 1 ) + 3 ( 2 + 1 ) + 4 ( 3 + 1 ) + .... + 10 ( 9 + 1 )

   = 1.0 + 1.2+2.3 + 3.4 + .... + 9.10 + 1 + 2 + 3 + ... + 10 

 Đặt B = 1.2 + 2.3 + ... + 9.10 

        C = 1 + 2 + 3 + ... + 10 

B = 1.2 + 2.3 + ... + 9.10

3B = 1.2.3 + 2.3.(4-1) + ... + 9.10.(11 - 8 )

    = 1.2.3 + 2.3.4 - 1.2.3 + ... - 8.9.10 + 9.10.11 

=> B = \(\frac{9.10.11}{3}=330\)

C  = \(\frac{10.11}{2}=55\)

=>A = B + C = 330 + 55 = 385 

A = 1.1 + 2.(1 + 1) + 3.(1 + 2) + ...+ 10.(1 +9)

A = 1 + 2 + 2.1 + 3 + 3.2 + ...+ 10 + 10.9

A = (1+2+3+...+ 10) + (1.2 + 2.3+ ...+ 9.10)

Tính 1+2+3+..+10 = (1 + 10).10 : 2 = 55

Tính B = 1.2 + 2.3 + ...+ 9.10 

3.B = 1.2.3 + 2.3. (4 - 1) + ...+ 9.10. (11- 8) = 1.2.3 + 2.3.4 - 1.2.3 + ...+ 9.10.11 - 8.9.10

3.B = (1.2.3 + 2.3.4 + ...+ 9.10.11) - (1.2.3 + ...+ 8.9.10) = 9.10.11 => B = 3.10.11 = 330

Vậy A = 55 + 330 = 385

A=1+2+2^2+2^3+2^4+2^5

2A=2+2^2+2^3+2^4+2^5+2^6

2A-A=(2+2^2+2^3+2^4+2^5+2^6)-(1+2+2^2+2^3+2^4+2^5)

A=2^6-1

A=64-1

A=63

B=10+12+14+....+2010

B=(2010+10).1001:2

B=2020.1001:2

B=2022020:2

B=1011010

\(A=2+2^2+2^3+...+2^{10}\)

\(2A=2^2+2^3+2^4+...+2^{11}\)

\(A=2A-A=2^{11}-2\)

\(B=\frac{72^3\cdot54^2}{108^4}=\frac{3^3\cdot3^3\cdot8^3\cdot3^2\cdot3^2\cdot3^2\cdot2^2}{3^4\cdot3^4\cdot3^4\cdot4^4}=\frac{3^{12}\cdot2^{11}}{3^{12}\cdot2^8}=2^3=8\)

a, A = 1+7+7²+7³+...+7¹⁰

7A = 7+7²+7³+7⁴+...+7¹¹

6A = 7A - A = 7¹¹ - 1

=> A = \(\frac{7^{11}-1}{6}\)

b, B = 1+3+3²+3³+...+3¹⁰⁰

3B = 3+3²+3³+3⁴+....+3¹⁰¹

2B = 3B - B = 3¹⁰¹ - 1

=> B = \(\frac{3^{101}-1}{2}\)

a) \(A=7^{11}--7\)

b) \(B=3^{101}-3\)

ta tách một số n như sau

\(n^2=n\cdot n=n\left(n-1+1\right)=n\left(n-1\right)+n\)

=> \(1^2=1\cdot0+1\)

\(2^2=2\cdot1+2\)

\(3^2=3\cdot2+3\)

.....

\(100^2=100\cdot99+100\)

=> \(B=\left(1\cdot0+2\cdot1+3\cdot2+\cdots+100\cdot99\right)+\left(1+2+\cdots+100\right)\)

nhân 3 vào ngoặc 1

=> \(=\frac{99.100.101}{3}=333300\) ( mik hơi lười vt bạn tự tra nha:v)

\(\Rightarrow B=333300+5050=338350\)

b) Áp dụng công thức tính tổng bình phương:

\(1^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) ( từ câu a)

tổng từ \(1^2\) đến \(200^2\) là:

\(\frac{200.201.401}{6}=2686700\)

tổng từ \(1^2\) đến \(100^2\) là: \(338350\)

=>C=\(2686700-338350=2348350\)

c) tách số như sau:

\(1\cdot3=1\cdot2+1\)

\(2\cdot4=2\cdot3+2\)

...

\(100\cdot102=100\cdot101+100\)

=> \(S=\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)+\left(1+2+3+\cdots+100\right)\)

=> \(S=\frac{100.101.102}{3}+\frac{100.101}{2}\)

=> \(S=343400+5050=348450\)

d) \(1\cdot100=1\cdot101-1^2\)

\(2\cdot99=2\cdot101-2^2\)

...

\(100\cdot1=100\cdot101-100^2\)

=>T=\(\left(1\cdot101+2\cdot101+\cdots+100\cdot101\right)-\left(1^2+2^2+\cdots+100^2\right)\)

\(T=101\left(1+2+3+\cdots+100\right)-B\) (B là tổng ở câu a) nha)

T=\(101\cdot5050-338350\)

\(T=171700\)

e) => \(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+\cdots+98\cdot99\cdot100\cdot4\)

\(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+\cdots+98\cdot99\cdot100\cdot\left(101-97\right)\)

\(4E=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+\cdots+98\cdot99\cdot100\cdot101-97\cdot98\cdot99\cdot100\)

=> \(4E=98\cdot99\cdot100\cdot101\)

=> \(E=24497550\)

\(A=2+2^2+2^3+...+2^{2010}\)

\(2A=2^2+2^3+2^4+...+2^{2011}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2011}\right)-\left(2+2^2+2^3+...+2^{2010}\right)\)

\(A=2^{1011}-2\)

Đặt A = 2¹ + 2² + ... + 2²⁰¹⁰

=> 2A = 2² + 2³ + ... + 2²⁰¹¹

=> 2A - A = ( 2² + 2³ + ... + 2²⁰¹¹ ) - ( 2¹ + 2² + ... + 2²⁰¹⁰ )

A = 2²⁰¹¹ - 2

VẬy A = 2²⁰¹¹ - 2

CÁC BẠN TRẢ LỜI NHANH NHÉ CHIỀU NAY PHẢI CÓ KẾT QUẢ

a/ Ta tính trường hợp tổng quát có n số hạng. Ta có:
+/ S1 = 1 + 2 + 3 + ....+n = \(\frac{n\left(n+1\right)}{2}\)
+/ S2 = 1.2 + 2.3 + 3.4 +...+ n(n+1)
3S2 = 1.2.3 + 2.3.3 + 3.4.3 +..+ n(n+1).3
3S2= 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +..+ n(n+1)(n+2 -(n-1))
3S2= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +.. - (n-1)n(n+1) + n(n+1)(n+2)
3S2= n(n+1)(n+2)
=> S2 = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Tính S = 1² + 2² + ...+ n²
Ta có: S2 - S1 = [1.2 + 2.3 + 3.4 +...+ n(n+1)]-(1 + 2 + 3 + ....+n)

=> S2 - S1=(1.2-1)+(2.3-2)+(3.4-3)+...+[n(n+1)-n]

=> S2 - S1=1+4+9+...+n²=1²+2²+3²+...+n²=S

Như vậy: S=S2-S1=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

=> \(S=n\left(n+1\right).\left(\frac{n+2}{3}-\frac{1}{2}\right)\)

=> \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Thay n=98 => \(S=\frac{98.99.197}{6}=318549\)

b/ 2014.2016=2014(2015+1)=2014+2014.2015=2014+2015(2015-1)=2014+2015²-2015=2015²-1<2015²

Vậy 2014.2016<2015²