Question

find the sum of series 3n^2-3n+2

Answer 1

 let S = ∑(3n² - 3n + 2)
         =3∑n² - 3∑n + 2∑1
  we know,
         ∑n² = n(n+1)(2n+1)/6 
         ∑n = n(n+1)/2 
         ∑1 = n   now use it above
S = 3n(n+1)(2n+1)/6 -3n(n+1)/2 +2n
   = n(n+1)(2n+1)/2 -3n(n+1)/2 + 4n/2 
   =n[(n+1)(2n+1)-3(n+1)+4]/2
   =n[(n+1)(2n+1 -3) + 4]/2
   =n[(n+1)(2n-2) + 4]/2
   =2n{(n+1)(n-1) + 2}/2 
   = n(n²+1)

Answer 2

∑ (1 to n)(3n2 - 3n + 2)

3 ∑ (1 to n) (n2 ) - 3 ∑ (1 to n) (n) + 2 ∑ (1 to n)

3 [ (1/6) n(n+1)(2n+1)] - 3 [(1/2)(n)(n+1)] + 2n

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

(1/2) n(n+1) [ 2n + 1 - 3] + 2n

n(n+1)(n - 1) + 2n

n ( n2 - 1 + 2 )

n (n2 + 1)