\(A=1+4+4^2+4^3+...+4^{50}\)

=>  \(4A=4+4^2+4^3+4^4+...+4^{51}\)

=>  \(4A-A=\left(4+4^2+4^3+...+4^{51}\right)-\left(1+4+4^2+...+4^{50}\right)\)

=>  \(3A=4^{51}-1\)

=>  \(A=\frac{4^{51}-1}{3}\)

\(4A=4.\left(1+4+4^2+...+4^{50}\right)\)

\(4A=4+4^2+4^3+...+4^{51}\)

\(4A-A=4+4^2+4^3+...+4^{51}-\left(1+4+4^2+...+4^{50}\right)\)

\(3A=4^{51}-1\)

\(A=\frac{4^{51}-1}{3}\)

Bài 5 :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

    \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{59}\)

     \(A=1-\frac{1}{50}\)

từ trên ta có : \(1-\frac{1}{50}< 1\)

\(\Rightarrow A< 1\)

a,Ta có :2A=2+2^2+2^3+...+2^2011

2A-A=2^2011-2^0=2^2011-1

b,Tính 4B làm tương tự A

a)2A=2(1+2.2+2.2²+...+2.2²⁰¹⁰)

=2.1+2.2+2.2²+...+2.2²⁰¹⁰

=2+2²+2³+...+2²⁰¹¹

2A-A=(2+2²+2³+...+2²⁰¹¹)-(1+2+2²+...+2²⁰¹⁰)

A=2²⁰¹⁰-1

Mình lộn 

S = \(\frac{\left(1+90\right)\times90\div2}{4}\)

S = \(\frac{4095}{4}\)

Tổng dãy đó là

  \(\left(\frac{90}{4}+\frac{1}{4}\right)\times\left[\left(\frac{90}{4}-\frac{1}{4}\right)\div\frac{1}{4}+1\right]\div2=\)\(=\frac{4095}{4}\)

     Đáp số : \(\frac{4095}{4}\)

ok tick luon khong thanh van de

Ai tick ủng hộ mik với

\(A=4^{2017}+4^{2016}+...+4^2+4+1\)

\(4A=4^{2018}+4^{2017}+...+4^3+4^2+4\)

\(4A-A=\left(4^{2018}+4^{2017}+...+4^3+4^2+4\right)-\left(4^{2017}+...+4+1\right)\)

\(3A=4^{2018}-1\)

\(A=\frac{4^{2018}-1}{3}\)

Chú ý: \(a^2-1=\left(a-1\right)\left(a+1\right)\)

Áp dụng:

\(A=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}...\frac{49.51}{50^2}=\frac{2.3.4^2.5^2...49^2.50.51}{3^2.4^2.5^2...50^2}=\frac{2.51}{3.50}=\frac{51}{75}\)