Question

ajutatima uuuuuuuuuuurrrrrrrrrrrrrgent!!!!!!​

Answer 1

Ajutați-mă* ;)

3+6+9+...+2010=3(1+2+3+...+670)=3*670*671/2

=2+2*3*670*671/2:671

=2+2*3*670*671/2*1/671

=2+2*3*670*1/2

=2+3*670

=2+2010

=2012

S=2^1+2^2+2^3+...+2^2009-2^2010

Înmulțesc toată relația cu 2.

2S=2^2+2^3+2^4+...+2^2009+2^2010-2^2011

Scad 2S-S:

2S-S=2^2+2^3+2^4+...+2^2009+2^2010-2^2011-(2^1+2^2+2^3+...+2^2009-2^2010)

S=2^2+2^3+2^4+...+2^2009+2^2010-2^2011-2^1-2^2-2^3-...-2^2009-2^2010

Termenii se reduc, rămânând ultimul și primul.

S=2^2011-2^1

S=2^1(2^2010-1)

2012+2(2^2010-1)=2[1006+(2^2010-1)]

Answer 2

Răspuns:

Explicație pas cu pas:

2+2×3(1+2+3+......+670):671+2+2²+2³+....+2^2009-2^2010=

=2+6×670×671/2:671+2×(2^2009-1)/(2-1)-2^2010=

=2+3×670+2×(2^2009-1)-2^2010=

=2+2010+2^2010-2-2^2010=2010