\(A=4+4^2+4^3+....+4^{99}+4^{100}\)

\(=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)

\(=4\cdot5+4^3\cdot5+...+4^{99}\cdot5\)

\(=5\left(4+4^3+...+4^{99}\right)\)

\(S=1\cdot2+2\cdot3+3\cdot4+...+2018\cdot2019\)

\(3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+2018\cdot2019\cdot3\)

\(3S=1\cdot2\cdot\left(3-0\right)+2\cdot3\left(4-1\right)+....+2018\cdot2019\left(2020-2017\right)\)

\(3S=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+....+2018\cdot2019\cdot2020-2017\cdot2018\cdot2019\)

\(3S=2018\cdot2019\cdot2020\)

\(S=\frac{2018\cdot2019\cdot2020}{3}\)

\(1\cdot2\cdot3+2\cdot3\cdot4+...+48\cdot49\cdot50\)

\(4P=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+48\cdot49\cdot50\cdot4\)

\(4P=1\cdot2\cdot3\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+....+48\cdot49\cdot50\left(51-47\right)\)

\(4P=1\cdot2\cdot3\cdot4-0\cdot1\cdot2\cdot3+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+....+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)

\(P=\frac{48\cdot49\cdot50\cdot51}{4}\)

\(Q=1^2+2^2+3^2+....+113^2\)

\(Q=1\left(2-1\right)+2\left(3-1\right)+....+133\left(134-1\right)\)

\(Q=\left(1\cdot2+2\cdot3+133\cdot134\right)-\left(1+2+3+...+133\right)\)

Áp dụng công thức cho nó nhanh:\(1\cdot2+2\cdot3+...+133\cdot134=\frac{133\cdot134\cdot135}{3}\)

\(1+2+3+...+133=\frac{133\cdot134}{2}\)

Đến đây đưa casio ra mak tính

A = 1.2 + 2.3 + 3.4 + ... + 98.99

A = 1.(1 + 1) + 2.(2 + 1) + 3.(3 + 1) + ... + 98.(98 + 1)

A = 1² + 1 + 2² + 2 + 3² + 3 + ... + 98².98

A = (1² + 2² + 3² + ... + 98²) + (1 + 2 + 3 + ... + 98)

A = (1² + 2² + 3² + ... + 98²) + 4851               (1)

B = 1² + 2² + 3² + ... + 98²              (2)

(1)(2) => A - B = 4851 ⋮ 4851

ta có: B = 1²  + 2² + 3² +...+98² = 1.1 +2.2+3.3+...+98.98

=> A-B = (1.2+2.3+3.4+4.5+...+98.99) - (1.1+2.2+3.3+...+98.98)

A-B = (1.2-1.1) + (2.3-2.2) + (3.4-3.3) + (4.5-4.4) + ...+ (98.99-98.98)

A-B = 1.(2-1) + 2.(3-2) +3.(4-3) + 4.(5-4) + ...+ 98.(99-98)

A-B = 1 +2+3+4+...+98

A-B = (1+98).98:2

A -B = 4851 chia hết cho 4851

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

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

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

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

=> \(E=24497550\)

\(A=\frac{\left(2018+1\right).2018}{2}=2037171\)

\(B=1.2+2.3+3.4+...+2018.2019\)

\(3B=1.2.3+2.3.3+3.4.3+...+2018.2019.3\)

\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2018.2019.\left(2020-2017\right)\)

\(3B=1.2.3+2.3.4-1.2.3+...+2018.2019.2020-2017.2018.2019\)

\(3B=2018.2019.2020\)

\(B=\frac{2018.2019.2020}{3}\)

\(B=2743390280\)

Chúc bạn học tốt ~ 

Ngu như con bò

vay sao chi

bài 1:

<=> \(x\left(x^2-\frac{9}{16}\right)=0\)

TH1:x=0

TH2: \(x^2-\frac{9}{16}=0\)

=> \(x^2=\frac{9}{16}\)

TH2a: \(\Rightarrow x=\frac34\)

\(TH2b:x=-\frac34\)

bài 2:

1) <=> \(2N=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\)

Áp dụng công thức: \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+1\right)}\) ta có:

\(2N=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\left(\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(2N=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)

\(2N=\frac{4949}{9900}\)

\(\Rightarrow N=\frac{4949}{19800}\)

2) <=> \(5N=5^2+5^3+5^4+\cdots+5^{101}\)

=> \(5N-N=\left(5^2+5^3+5^4+\cdots+5^{101}\right)-\left(5+5^2+5^3+\cdots+5^{99}+5^{100}\right)\)

\(4N=5^{101}-5\)

=> \(N=\frac{\left(5^{101}-5\right)}{4}\)

Bài 1: 

a)CMR: ab + ba chia hết cho 11 

Theo đề bài ta có: ab + ba = (10a + b) + (10b + a)

                                         = 11a + 11b chia hết cho 11                                                                                                                                                                                                                                                                                                              b)CMR: abc - cba chia hết cho 99

Theo đề bài ta có: abc - cba = (100a - 10b - c) + (100c - 10b - a)

                                         = 99a - 99c chia hết cho 99

Bài 2

  A= (3²¹ + 3²² + 3²³⁾ + ... + (3²⁷ + 3²⁸ + 3²⁹⁾                                                                                                                                                                               A= 3²¹.(1 + 3 + 3²) + ... + 3²⁷. (1 + 3 + 3²)                                          

  A=3²¹ . 13 + ... + 3²⁷ . 13  

  A= 13 . (3²¹ + ... + 3²⁷) chia hết cho 13