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