Question 1131970
<pre>
{{{matrix(1,3,sum((n^2),n=0,k),""="",expr((k(k+1)(2k+1))/6))}}}

k=1

{{{matrix(1,13,sum((n^2),n=0,1),""="",1^2,""="",1,""="",expr((1(1+1)(2*1+1))/6),
""="",expr((1(2)(3))/6),""="",6/6,""="",1

)}}}

k=2

{{{matrix(1,15,sum((n^2),n=0,2),""="",1^2+2^2,""="",1+4,""="",5,""="",expr((2(2+1)(2*2+1))/6),
""="",expr((2(3)(5))/6),""="",30/6,""="",5

)}}}

k=3

{{{matrix(1,15,sum((n^2),n=0,3),""="",1^2+2^2+3^2,""="",1+4+9,""="",14,""="",expr((3(3+1)(2*3+1))/6),
""="",expr((3(4)(7))/6),""="",84/6,""="",14

)}}}

---------------------------------------------

{{{matrix(1,7,
sum((2m^2+3m-4),n=0,5),
""="",
sum((2m^2),m=0,5)+sum((3m),m=0,5)+sum((-4),m=0,5),
""="",
2sum((m^2),m=0,5)+3sum((m),m=0,5)+sum(((-4)*1^""),m=0,5),
""="",
2sum((m^2),m=0,5)+3sum((m),m=0,5)-4sum((1),m=0,5)

)}}}

The preceding two steps involved (A) taking out a constant coefficient,
and (B) introducing a multiplier of 1 so we would have a constant
coefficient of -4 to take out.  But from here we need two other summation
formulas, which are:

        {{{matrix(1,3,sum((n),n=0,k),""="",expr((k(k+1))/2))}}} and {{{matrix(1,3,sum((1),n=0,k),""="",k+1))}}}

{{{matrix(1,6,
sum((2m^2+3m-4),n=0,5),
""="",
2sum((m^2),m=0,5)+3sum((m),m=0,5)-4sum((1),m=0,5),
""="",
2*expr((5(5+1)(2*5+1))/6)+3*expr((5(5+1))/2)-4(5+1), ""=""

)}}}

{{{matrix(1,9,

2*expr((5(5+1)(2*5+1))/6)+3*expr((5(5+1))/2)-4(5+1), ""="",

2*expr((5(6)(11))/6)+3*expr((5(6))/2)-4(6), ""="",

2*55+3*15-24, ""="",110+45-24,""="", 131

)}}}

Edwin</pre>