Question

B=\(\dfrac{\text{𝟏+(𝟏+𝟐)+(𝟏+𝟐+𝟑)+⋯+(𝟏+𝟐+⋯.+𝟐𝟎𝟏𝟐)}}{\text{𝟏 𝒙 𝟐𝟎𝟏𝟐+𝟐 𝒙 𝟐𝟎𝟏𝟏+𝟑 𝒙 𝟐𝟎𝟏𝟎+⋯+𝟐𝟎𝟏𝟐 𝒙 1}}\)

Answer 1

B=\(\dfrac{\text{1+(1+2)+(1+2+3)+⋯+(1+2+⋯.+2012)}}{\text{2012 × 1 + 2 × 2011 + . . . . . . . . + 1 × 2012}}\)

=\(\dfrac{\text{1 + 1 + 2 + 1 + 2 + 3 + . . . . . . . . + 1 + 2 + 3 + . . . . . . + 2012}}{\text{2012 × 1 + 2011 × 2 + . . . . . . . . . + 1 × 2012}}\)

=\(\dfrac{\text{( 1 + 1 + 1 + . . . . . + 1 ) + ( 2 + 2 + 2 + 2 + . . . . . . + 2 ) + . . . . . . . . + 2012 }}{\text{1 × 2012 + 2 × 2011 + . . . . . . . . . . + 2012 × 1}}\)

=\(\dfrac{\text{1 × 2012 + 2 × 2011 + . . . . . . . + 2012 × 1}}{\text{1 × 2012 + 2 × 2011 + . . . . . . . + 2012 × 1 }}\)=1