Ta có:

\(C= 4+44+444+......+4444444444\)

\(C= 4.(10.1+9.10+8.100+7.1000+...+1.1000000000\)

\(C= 4.(100+90+800+7000+60000+500000+4000000+30000000+200000000+1000000000)\)

\(C=4.12345678900\)

\(C=4938271600\)

Tương tự.

(chỉnh đề)

A=\(-1+2-3-4-5+6-7-8-9+...-2021-2022+2023-2024\)

=\(\left(-1-2024\right)+\left(2+2023\right)+\left(-3-2022\right)+\left(-4-2021\right)+\left(-5-2020\right)+\left(6+2019\right)-\left(-7-2018\right)+\left(-8-2017\right)+\left(-9-2016\right)+...+\left(1010+1015\right)+\left(-1011-1014\right)+\left(-1012-1013\right)\)=\(-2025+2025-2025-2025-2025+2025-2025-2025-2025+...+2025-2025-2025\)=253.2025-1771.2025=-3 073 950.

B=\(1.3.5+3.5.7+5.7.9+7.9.11+...+99.101.103\)

8B=\(1.3.5.8+3.5.7.8+5.7.9.8+7.9.11.8+...+99.101.103.8\)

8B=\(1.3.5.\left[7-\left(-1\right)\right]+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+7.9.11.\left(13-5\right)+...+99.101.103.\left(105-97\right)\)8B=\(3.5+3.5.7+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+7.9.11.13-5.7.9.11+...+99.101.103.105-97.99.101.103\)

B=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)

- Đặt C₁=\(1.3+3.5+...+197.199\).

6C₁=\(1.3.6+3.5.6+...+197.199.6\)

6C₁=\(1.3.\left[5-\left(-1\right)\right]+3.5.\left(7-1\right)+...+197.199.\left(201-195\right)\)

6C₁=\(3+3.5+3.5.7-3.5+...+197.199.201-195.197.199\)
C₁=\(\dfrac{3+197.199.201}{6}=1313301\)

- Đặt C₂=\(2.4+4.6+...+198.200\)

6C₂=\(2.4.6+4.6.6+...+198.200.6\)

6C₂=\(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+198.200.\left(202-196\right)\)

C₂=\(\dfrac{198.200.202}{6}=1333200\)

=>C₁+C₂=C=1313301+1333200=2646501

c=94/285

mình cũng ko hiểu cách giải này cho lắm nên mình làm thế thôi

Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)

Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)

\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)

Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với  \(n\in N\)

Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)

Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)

Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)