Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
NHỚ TK MK NHA BN!
Ta có:\(n^2-n=n\left(n+1\right)\Leftrightarrow n^2=n\left(n+1\right)+n\)
Áp dụng \(A=1^2+2^2+3^2+..+9^2=\left(1.0+1\right)+\left(2.1+2\right)+\left(3.2+3\right)+...+\left(9.8+9\right)\)
\(=1.2+2.3+3.4+...+8.9+1+2+3+...+9\)
Xét \(a=1.2+2.3+...+8.9\)
\(3a=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+8.9\left(10-7\right)\)
\(3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+7.8.9-8.9.10\)
\(3a=8.9.10\Rightarrow a=\frac{8.9.10}{3}=240\)
\(\Leftrightarrow A=240+1+2+...+9\)
\(=240+\frac{\left(1+9\right)9}{2}=240+45=285\)
Vậy A=285
a: \(3\sqrt{200}=3\cdot10\sqrt{2}=30\sqrt{2}\)
b: \(-5\sqrt{50a^2b^2}=-5\cdot5\sqrt{2a^2b^2}\)
\(=-25\cdot\left|ab\right|\cdot\sqrt{5}\)
c: \(-\sqrt{75a^2b^3}\)
\(=-\sqrt{25a^2b^2\cdot3b}=-5\left|ab\right|\cdot\sqrt{3b}\)
\(A=2^1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\cdot\left(2+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow2A=2^2+2^3+...+2^{11}\)
\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{11}\right)-\left(2+2^2+...2^{10}\right)\)
\(\Rightarrow A=2^{11}-2\)
\(B=3^1+3^2+...+3^{100}\)
\(\Rightarrow3B=3\cdot\left(3+3^2+...+3^{100}\right)\)
\(\Rightarrow3B=3^2+3^3+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-3\)
\(\Rightarrow B=\dfrac{3^{101}-3}{2}\)
phần B thiếu 3 mũ 3 ak