Question

Pls help urgent....
Question from some special series and AP

Answer 1

Answer:

n(n+1)( 3n² + n + 2) /12

Step-by-step explanation:

First Term = r

Common Difference = 2r - 1

Sum of r terms = (r/2) ( r  + r + (r-1)(2r-1))

= (r/2) ( 2r  + 2r² - 3r + 1)

= (r/2)(2r² - r + 1)

= r³  - r²/2  + r/2

Sum = V₁ + V₂ +...........................+ Vₙ

=  ∑ (r³  - r²/2  + r/2)

=  n²(n+1)²/4   - n(n+1)(2n+1)/12   +  n(n+1)/4

= ((n)(n+ 1)/ 12) ( 3n(n+1) - (2n+1) + 3)

= ((n)(n+ 1)/ 12)( 3n² + 3n - 2n - 1 + 3)

=  ((n)(n+ 1)/ 12)( 3n² + n + 2)

=  ((n)(n+ 1)/ 12)( 3n² + n + 2)

= n(n+1)( 3n² + n + 2) /12