Question

[nguyen thi thien lam]

1)tính

S=2^2+4^2+6^2...+20^2 biet 1^2+2^2+3^2+...+10^2=385

2)tính

A=2^0+2^1+2^3+...+2^100

Answer 1

[Lightning Farron]

Bài 1:

\(S=2^2+4^2+6^2+...+20^2\)

\(=\left(1\cdot2\right)^2+\left(2\cdot2\right)^2+\left(2\cdot3\right)^2+...+\left(2\cdot10\right)^2\)

\(=1\cdot2^2+2^2\cdot2^2+2^2\cdot3^2+...+2^2\cdot10^2\)

\(=2^2\left(1+2^2+3^2+...+10^2\right)\)

\(=4\cdot385=1540\)

Bài 2:

\(A=2^0+2^1+2^2+...+2^{100}\)

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

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

\(2A=2+2^2+2^3+...+2^{101}\)

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

\(A=2^{101}-1\)

Answer 2

[Phương Trâm]

Giải:

\(1.\) \(S=2^2+4^2+6^2+....+20^2\)

\(2^2=\left(1.2\right)^2\)

\(4^2=\left(2.2\right)^2\)

\(...\)

Vế dưới \(= \left(1.2\right)^2 + \left(2.2\right)^2 + ...+ \left(9.2\right)^2+ \left(10.2\right)^2\)

\(= 2^2.(1^2 + 2^2 + 3^2 + ...+ 9^2 + 10^2) \)

\(= 4. 385\)

\(= 1540\)

\(2.\)

\( 2A = 2^1 + 2^2 + 2^3 + 2^4 +...+\)\(2^{2011}\)

\(2A - A = ( 2^1 + 2^2 + 2^3+ 2^4 +...+ 2^{2011} ) - ( 1 + 2^2 + 2^3 +...+ 2^{2010} ) \)

\(\Rightarrow A = 2^{2011} - 1\)