cau hoi nay ai chang lam dc

học vs hành, bỏ đi, bỏ đi, ra cái đề câu mô cụng hỏi rk hk mần chi

Ta có công thức:

\(\frac{n^2\times\left(n+1\right)^2}{4}\)

\(\Rightarrow\frac{99^2\times100^2}{4}=24502500\)

Chúc bạn học tốt.

M=3+3²+3³+...+3ⁿ

=>3M=3²+3³+3⁴+...+3ⁿ⁺¹

=>3M-M=3ⁿ⁺¹-3

=>2M=3ⁿ⁺¹-3

=>M=\(\frac{3^{n+1}-3}{2}\)

\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)

=>3N\(=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\)

=>3N-N=\(1-\frac{1}{3^n}\)

=>2N=\(1-\frac{1}{3^n}\Rightarrow N=\frac{1-\frac{1}{3^n}}{2}\)

help me

\(A=\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow3A=1+\frac{1}{3^0}+\frac{1}{3^1}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)

\(\Rightarrow A=\frac{3^{2005}-1}{3^{2005}.2}\)

cảm ơn thành đạt 

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(2A=1-\frac{1}{3^8}\)

\(A=\frac{\left(\frac{6560}{6561}\right)}{2}=\frac{3280}{6561}\)

\(A=\frac{3280}{6561}\)

Ta có:

1³ + 2³ + ... + n³ = [1+2+3+...+n]²

Áp dụng vào bài toán:

1³ + 2³ + 3³ + 4³ + 5³ = [1+2+3+4+5]² = 15²

Bài 1 :Áp dụng

\(S100=1+a+a^2+...+a^{100}=\frac{a^{101}-1}{a-1}\)

Với : \(a=-2\),ta được 

\(S=1-2+2^2-2^3+...+2^{100}\)

    \(=\frac{\left(-2\right)^{100}-1}{-2-1}=\frac{-2^{101}-1}{-3}=\frac{2^{101}+1}{3}\)

      \(T=3-3^2+3^3-...+3^{1998}-3^{2000}\)

          \(=3\left(1-3+3^2-3^3+...+3^{1998}-3^{1999}\right)\)

           \(=3.\frac{\left(-3\right)^{2000}-1}{-3-1}=3.\frac{3^{2000}-1}{-4}\)

          \(=\frac{3.\left(1-3^{2000}\right)}{4}\)

Chúc bạn học tót ( -_- )

\(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2=\frac{n^2\left(n+1\right)^2}{4}\)

thank you