A = \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+...+2004}\)+ \(\dfrac{2}{2025}\)

A = \(\dfrac{1}{\left(1+3\right).3:2}\)+\(\dfrac{1}{\left(4+1\right).4:2}\)+...+ \(\dfrac{1}{\left(2024+1\right).2024:2}\)+\(\dfrac{2}{2025}\)

A = \(\dfrac{2}{3.4}\)+\(\dfrac{2}{4.5}\)+...+\(\dfrac{2}{2024.2025}\)+ \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{2024.2025}\)) + \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)

A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)

A = \(\dfrac{2}{3}\) - \(\dfrac{2}{2025}\) + \(\dfrac{2}{2025}\)

A  = \(\dfrac{2}{3}\) 

Cc

\(2^3+4^3+6^3+....+18^3=\left(2.1\right)^3+\left(2.2\right)^3+\left(2.3\right)^3+...+\left(2.9\right)^3\)

\(=2^3.1^3+2^3.3^3+....+2^3.9^3=2^3\left(1^3+2^3+....+9^3\right)\)

\(=2^3.2025=8.2025=16200\)

Ta có :

2³= 2.2.2=1.2.1.2.1.2  và 1³=1.1.1

So sánh : 1.2.1.2.1.2 / 1.1.1 = 2.2.2 = 8

4³= 4.4.4 = 2.2.2.2.2.2 và 2³=2.2.2

So sánh : 2.2.2.2.2.2 / 2.2.2 = 2.2.2 = 8

Tương tự các cặp số lũy thừa khác.

...18³=18.18.18=9.2.9.2.9.2 và 9³= 9.9.9 

So sánh : 9.2.9.2.9.2 / 9.9.9 = 2.2.2 = 8.

Kết luận : 

1³+2³+3³+...+9³ / 2³+4³+6³+...+18³ = 1/8.

Vậy 2³+4³+6³+...+18³= 2025.8 = 16200.

                                                                 Đáp số : 16200.

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}].[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}]}\)

=\(\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

=\(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Áp dụng ta có S=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)

Ta có công thức tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2025\sqrt{2024}+2024\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)

B=2^3(1^3+2^3+...+9^3)

=8*2025=16200

Đỉnh cao đấy. Nhanh gọn lẹ luôn

a: Ta có: \(A=2+2^2+2^3+\cdots+2^{2025}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{2026}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{2026}-2-2^2-\cdots-2^{2025}\)

=>\(A=2^{2026}-2\)

b:Sửa đề: \(B=1+5+5^2+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

c: Ta có: \(C=3+3^2+3^3+\ldots+3^{1000}\)

=>\(3C=3^2+3^3+3^4+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

1416

Khó quá thì lên hoc24.vn

\(A=2^3\left(1^3+2^3+3^3+...+9^3\right)\)

\(=8\cdot2025=16200\)

ta có : 1^3+2^3+...+9^3=2025

=>    2.(1^3+4^3+6^3+.....+18^3)=2025.2

=>    2^3+4^3+...+18^3 =4050

 Vậy 2^3+4^3+...+18^3=4050

Ta có : 2^3 + 4^3 + 6^3 + ... + 18 ^3

= ( 1.2 )^3 + ( 2.2 ) ^3 + ( 2 .3 ) ^3 + .... + ( 2 .9 ) ^3

= 1^3 . 2^3 + 2^3 . 2^3 + 2^3 . 3^3 + ... + 2^3 . 9^3

= 2^3 . ( 1^3  + 2^3 + 3^3 + ... + 9^3 )

= 8 . 2025 ( vì 1^3  + 2^3 + 3^3 + ... + 9^3 = 2025)

= 16200