Tính A = : \(1^3+2^3+3^3+...+100^3\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,S=1+3+3^2+....+3^{100}.\)
\(\Rightarrow3S=3+3^2+...+3^{101}\)
\(\Rightarrow3S-S=\left(3+3^2+...+3^{101}\right)-\left(1+3+....+3^{100}\right)\)
\(\Rightarrow2S=3^{101}-1\)
\(\Rightarrow S=\frac{3^{101}-1}{2}\)
\(b,A=1+3^2+3^4+...+3^{100}\)
\(\Rightarrow3^2A=3^2+3^4+...+3^{102}\)
\(\Rightarrow9A-A=\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+....+3^{100}\right)\)
\(\Rightarrow8A=3^{102}-1\)
\(\Rightarrow A=\frac{3^{102}-1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2+22+…+2100
2A=2(1+2+22+…+2100)
2A=2+22+…+2101
2A-A = A = 2+22+…+2101-(1+2+22+…+2100)
A = 2+22+…+2101-1-2-22-…-2100
A = (2-2)+(22-22)+…+(2100-2100)+2101-1
A = 0+0+…+0+2101-1
A = 2101-1
B=3-32+33-34+…+299-3100
3B = 3(3-32+33-34+…+299-3100)
3B = 32-33+34-…-299+3100-3101
3B+B = 4B = 3-32+33-34+…+299-3100
4B =(3-32+33-34+…+299-3100)+(32-33+34-…-299+3100-3101)
4B =3-32+33-34+…+299-3100+32-33+34-…-299+3100-3101
4B =3+(32-32)+(33-33)+(34-34)+…+(299-299)+(3100-3100)-3101
4B =3+0+0+0+....+0-3101
4B =3-3101
B = (3-3101)/4
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số hạng tổng quát trong tổng n3
Nhận xét: n3 - n = n(n2 - 1) = (n - 1).n.(n + 1) => n3 = (n - 1).n.(n + 1) + n. Áp dụng ta có:
13 = 0 + 1
23 = 1.2.3 + 2
33 = 2.3.4 + 3
....
1003 = 99.100.101 + 100
=> A = (1 + 2+3+...+100) + (1.2.3 + 2.3.4 + ...+ 99.100.101)
Tính B = 1.2.3 + 2.3.4 + ...+ 99.100.101
4.B = 1.2.3.4 + 2.3.4.(5 - 1) + ...+ 99.100.101.(102 - 98)
4.B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + ...+ 99.100.101.102 - 98.99.100.101 = 99.100.101.102
=> 4.B - B = 99.100.101.102 => B = 99.100.101.102 : 3
Tính C = 1 + 2+3 + ...+ 100 = (1+100).100 : 2 = ...
Vậy A = C + B = ..
ta xet :
1^3=0+1
2^3=1*2*3+2
3^3=2*3*4+3
....................
100^3=99*100*101+100
=>A=(1*2*3+2*3*4+....+99*100*101)+(1+2+3+..+100)
=>A=A1+A2
ta co
A1=1*2*3+2*3*4+....+99*100*101
4A1=1*2*3*4+2*3*4*(5-1)+...+99*100*101*(102-98)
4A1=1*2*3*4+2*3*4*5-1*2*3*4+....+99*100*101*102-98*99*100*101
4A1=(99*100*100)/4
A1=249975
ta co
A2=1+2+3+4+....+100
A2=(100+1)*100/2
A2=5050
=>A=A1+A2
=>A=249975+5050=255025
Vay A=255025