A = 1.2 + 2.3 + ... + 99.100

3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

3A = 999900

A = 333300

C = 1.2.3 + 2.3.4 + ... + 49.50.51

4C = 1.2.3.4 + 2.3.4.(4-1) + ... + 49.50.51.(52-48)

4c = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ... + 49.50.51.52 - 48.49.50.51

4C = 49.50.51.52

4C = 6497400

C = 1624350

Ta có :

a=1.2+2.3+3.4+...+99.100

3a=1.2.3+2.3.3+3.4.3+...+99.100.3

3a=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3a=99.100.101

a=\(\frac{99.100.101}{3}\)

a=333300

Tính c làm tương tự

\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)

Suy ra \(E=A-B=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

\(\frac{14949}{20200}\)

A) \(A=1+4+4^2+4^3+...+4^{26}\)

\(\Rightarrow4A=4+4^2+4^3+4^4+...+4^{27}\)

\(\Rightarrow4A-A=4^{27}-1\)

\(3A=4^{27}-1\)

\(A=\frac{4^{27}-1}{3}\)

B) \(B=3+3^3+3^5+3^7+...+3^{21}\)

\(\Rightarrow3^2B=3^3+3^5+3^7+3^9+...+3^{23}\)

\(\Rightarrow3^2B-B=3^{23}-3\)

\(8B=3^{23}-3\)

\(B=\frac{3^{23}-3}{8}\)

C) \(M=1.2+2.3+3.4+...+49.50\)

\(\Rightarrow3M=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(3M=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(3M=\left(1.2.3+2.3.4+3.4.5+...+49.50.51\right)-\left(1.2.3+3.4.5+...+48.49.50\right)\)

\(3M=49.50.51\)

\(M=\left(49.50.51\right):3\)

\(M=41650\)

D) \(N=1.2.3+2.3.4+3.4.5+...+49.50.51\)

\(\Rightarrow4N=1.2.3.4+2.3.4.4+3.4.5.4+...+49.50.51.4\)

\(4N=1.2.3.\left(4-0\right)+2.3.4\left(5-1\right)+3.4.5.\left(6-2\right)+...+49.50.51.\left(52-48\right)\)

RỒI BN LÀM GIỐNG NHƯ MK Ở PHẦN C THÌ NÓ SẼ RA!

CHÚC BN HỌC TỐT!!!!

c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4

==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)

==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11

==> 4C= 8.9.10.11=7920

==> C= 7920 :4=1980

a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3

               3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)

               3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)

               3A= 99.100.101 - 0.1.2

               3A= 999900 - 0

               3A= 999900

    ==> A= 999900 : 3

   ==> A= 333300

kết bạn kiểu gì

a)1+3+5+7+9+...+x=1600

=>[(x-1):2+1].(x+1)/2=1600

=>(1/2.x-1/2+1).(x+1)=1600:1/2

=>(1/2.x-1/2+1).(x+1)=3200

=>(x+1)².1/2=3200

=>(x+1)²       =3200:1/2

=>(x+1)²=6400

=>x+1=80

=>x=80-1=79

mik nham nhe bai 2

c, (x+1)+(x+2)+...+(x+100)=5750

\(A = 1.2+2.3+3.4+4.5+...+99.100\)

\(3A= 1.2.3+2.3.3+3.4.3+4.5.3+\)\(...+\)

\(99.100.3\)

\(3A = 1.2.3+2.3.(4-1)+3.4. (5-2)+\)

\(4.5. (6-3)+...+99.100. (101-98)\)

\(3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+\)

\(4.5.6-3.4.5+...+99.100.101-98.99.100\)

\(3A = 99 .100 .101\)

\(A = 99 .100 . 101 ÷ 3 \)

\(A = 333300\)

A = 1.2 + 2.3 + 3.4 + ....... + 99.100
3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 99 . 100 . 3
3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +.... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 99 . 100 . 101 - 98 . 99 . 100
3A = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99 . 100 . 101
3A = 99 . 100 . 101 = 999900
A = 999900 : 3 = 343400

# Học tốt☘️#

3F= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)] 
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)] 
=n(n+1)(n+2) 
=>F 

H=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

=> 4H=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)((n+3)-(n-1))

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)

=n(n+1)(n+2)(n+3)