Equivalent Expressions

  (b)  1/6k(k + 1)(2k + 1)                (a)  1/6k(k + 1)(2k + 1) + (k + 1)^2  

Expanding b with the substitution, we have

(1 / 6) (k + 1) ( [k + 1]  + 1) (2 [k + 1] + 1)  =

(1 /6) ( k + 1) (k + 2) (2k + 3)   =

(1/6) ( k^2 + 3k + 2)(2k + 3)  =

(1 / 6) (2k^3 + 6k^2 + 4k + 3k^2 + 9k + 6)  =

(1 /6) (2k^3 + 9k^2 + 13k + 6)      (1)

Expanding a, we have

1/6k(k + 1)(2k + 1) + (k + 1)^2   =

(1 / 6) (k^2 + k)(2k + 1)  + k^2 + 2k + 1  =

(1/6) ( 2k^3 + 2k^2 + k^2 + k) + k^2 + 2k + 1 =

(1/6) (2k^3 + 2k^2 + k^2  + k)   + (1/6)(6k^2 + 12k + 6)  =

(1 / 6) ( 2k^3 + 9k^2 + 13k + 6)       (2)

And since (1)  = (2).....the expressions are equivalent

Thank you so much!

You're welcome, Shades.....note that we could have also factored  (a) as,

1/6k(k + 1)(2k + 1) + (k + 1)^2   =

(1/6)[k + 1] [ k[2k + 1]  + [k + 1] [k + 1] =

(1/6) [k + 1] [ k[2k + 1] + 6 (k +1) ]  =

(1/6) [ k + 1] [ 2k^2 + 7k + 6]  =

(1/6) [k + 1] [ k + 2 ] [ 2k + 3]

And this would have been  equal  to  the substitution in (b), as follows ::

(1 / 6) (k + 1) ( [k + 1]  + 1) (2 [k + 1] + 1)  =

(1 /6) [ k + 1] [k + 2] [2k + 3]   =