I'm rusty too, so I'll have to derive it from scratch.
First we'll find a formula for 

S(k) = 

then your summation is

S(a+n-1) - S(a-1)

{S(k)} = 1², 1²+2², 1²+2²+3², ··· , 1²+2²+3²+···+k²

Write out some of the terms

1, 5, 14, 30, 55, 91, 140

Make a difference table, i.e, list the terms
vertically in a column placing the difference 
between each pair of successive terms between 
them to the right.  Then do the same to the 
second column, until you get a column which all
contain the same number:

  1
     4
  5     5
     9     2
 14     7 
    16     2
 30     9
    25     2
 55    11 
    36     2
 91    13
    49
140

It took three difference columns to get a column of
all 2's, so we will see if a third degree polynomial
in k is possible for the formula:

S(k) = Ak³ + Bk² + Ck + D

Then substituting  

 1 = A(1)³ + B(1)² + C(1) + D
 5 = A(2)³ + B(2)² + C(2) + D
14 = A(3)³ + B(3)² + C(3) + D
30 = A(4)³ + B(4)² + C(4) + D

Giving us this system of 4 equations in 4 unknowns

  A +   B +  C + D = 1
 8A +  4B + 2C + D = 5 
27A +  9B + 3C + D = 14
64A + 16B + 4C + D = 30

Solve this system and we get

A = 1/3, B = 1/2, C = 1/6, D = 0

So if we are right our formula is

S(k) = (1/3)k³ + (1/2)k² + (1/6)k

S(k) = (2k³ + 3k² + k)/6

S(k) = k(2k² + 3k + 1)/6

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

This can be proved correct by induction.
It is true for k=1, so if we add (k+1)²
to both sides

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

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

= [(k+1)/6][2k²+7k+6] = [(k+1)/6][(k+2)(2k+3)]

= (k+1)(k+2)(2k+3)/6 which equals S(k+1)

so we have the right formula.

Now your problem is to find 

S(a+n-1) - S(a-1)

(a+n-1)(a+n-1+1)(2(a+n-1)+1)/6 -
                      (a-1)(a-1+1)(2(a-1)+1)/6

(a+n-1)(a+n)(2a+2n-1)/6 - (a-1)(a)(2a-1)/6 =

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

I'll let you simplify that if you like.
I'm too tired! :-)

Edwin
AnlytcPhil@aol.com