mình biết nội quy rồi nên đưng đăng nội quy

ai chơi bang bang 2 kết bạn với mình

mình có nick có 54k vàng đang góp mua pika 

ai kết bạn mình cho

(1^2+2^2+3^+....+10^2).(65.111-13.15.17)

=(1^2+2^2+3^2+...+10^2).(65.111-13.555)

=(1^2+2^2+3^2+...+10^2).(65.111+13.5.111)

=(1^2+2^2+3^2+...+10^2).(65.111+65.111)

=(1^2+2^2+3^2+...+10^2).[111.(65-65)

=(1^2+2^2+3^2+...+10^2).(111.0)

=(1^2+2^2+3^2+...+10^2).0

=0

Đặt A = \(1+2+2^2+2^3+2^4+....+2^{100}\)

2A = \(2\left(1+2+2^2+2^3+2^4+....+2^{100}\right)\)

= \(2+2^2+2^3+2^4+2^5+...+2^{101}\)

2A - A = \(\left(2+2^2+2^3+2^4+2^5+....+2^{101}\right)-\left(1+2^2+2^3+2^4+...+2^{100}\right)\)

= \(2^{101}-1\)

Nếu bạn bt lm r thì ko nên ra câu hỏi nx đâu .

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)

(1^2+2^2+...+2016^2).(65.111-13.15.17)

=(1^1+2^2+...+2016^2).0

=0

Có : \(S=1+2+2^2+2^3+....+2^{99}\)

\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)

\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)

\(\Rightarrow S=2^{100}-1< 2^{100}\)

Vậy \(S< 2^{100}\)

 S=1+2+2²+2³+....+2⁹⁹

⇒2S=2+2²+2³+....+2¹⁰⁰

⇒2S−S=2¹⁰⁰-1

S=2¹⁰⁰-1

vì 2¹⁰⁰ -1<2¹⁰⁰

⇒S<2¹⁰⁰

\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+...+2^{100}\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+6.2^2+...+6.2^{98}\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)