\(S=1+2+2^2+2^3+...+2^{62}+2^{63}\)

\(2S=2\left(1+2+2^2+2^3+...+2^{62}+2^{63}\right)\)

\(2S=2+2^2+2^3+2^4+...+2^{63}+2^{64}\)

\(2S-S=\left(2+2^2+2^3+2^4+...+2^{63}+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{62}+2^{63}\right)\)

\(S=2^{64}-1\)

Bài toán làm theo kiểu 2.S là được nếu là 3^x thì sử dụng 3.S. Tương tự như vậy

Ta có: 1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³

\(\Rightarrow\) 2.(1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³) trừ (1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³) = 1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³

= (2 + 2² + 2³ + 2⁴ +...+ 2⁶³ + 2⁶⁴) trừ (1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³)

(Sử dụng phương pháp chịt tiêu: (là thế này nè)

 (2 + 2² + 2³ + 2⁴ +...+ 2⁶³ + 2⁶⁴) trừ (1 + 2 + 2² + 2³ +...+ 2⁶² + 2⁶³)

Còn lại 2⁶⁴ trừ 1)

= 2⁶⁴ trừ 1

Vậy S = 2⁶⁴ trừ 1

S = 1 - 2 + 2² - 2³ + ....... + 2²⁰²⁰

2S = 2(1 - 2 + 2² - 2³ + ....... + 2²⁰²⁰)

2S = 2 - 2² + 2³ - 2⁴ + ....... + 2²⁰²¹

S = (2 - 2² + 2³ - 2⁴ + ....... + 2²⁰²¹) - (1 - 2 + 2² - 2³ + ....... + 2²⁰²⁰)

S = 2²⁰²¹ - 1

3S = 3(2²⁰²¹ - 1)

3S - 2²⁰²¹ = 3(2²⁰²¹ - 1) - 2²⁰²¹

3S - 2²⁰²¹ = 3.2²⁰²¹ - 3 - 2²⁰²¹

➤ 3S - 2²⁰²¹ = 2²⁰²¹ . 2 - 3

sao ko dung f(x) ma viet

\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^9+2^{10}\)

a=\(\left(2+2^2\right)+2^2.\left(2+2^2\right)+..+2^8\left(2+2^2\right)\)

a=\(\left(2+2^2\right).\left(1+2^2+..+2^8\right)\)

a=\(6.\left(1+2^2+2^4+2^6+2^8\right)\)

chia het cho 3

S = - 504

olm duyệt đi

tinh the nao ma ra nhu the day

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\cdots+\frac{1}{\left(2n\right)^2}\)

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

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)

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

=>\(\frac14\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)<\frac14\)

=>\(A<\frac14\) (ĐPCM)

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)

A=21+22+23+...+22010

=(21+22)+(23+24)+...+(22009+22010)=(21+22)+(23+24)+...+(22009+22010)

=2(1+2)+23(1+2)+...+22009(1+2)=2(1+2)+23(1+2)+...+22009(1+2)

=3(2+23+...+22009)⋮3=3(2+23+...+22009)⋮3

�=21+22+23+...+22010A=21+22+23+...+2...

 \(10^6\) tận cùng là 0 \(=>10^6+2\) tận cùng là 2 \(=>10^6+2\) chia hết cho 2