Question 1008316
.
Please solve:
Sn=a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+....+[(a+(n-1)d]^2
Sn=a²+(a+d)²+(a+2d)²+(a+3d)²+....+[(a+(n-1)d]²
----------------------------------------------------

<pre>
Let us consider one typical term {{{(a+kd)^2}}}. It is
{{{a^2}}} + {{{2*k*d}}} + {{{k^2*d^2}}}.

We need to sum up n such terms/trinomials from k = 0 to k = n-1.

By combining the first addends of these trinomials, you will get {{{n*a^2}}}, right?

By combining the second addends of these trinomials, you will get {{{2d*(0 + 1 + 2 + 3 + ellipsis + (n-1))}}}.

If you know it, this sum {{{1 + 2 + 3 + ellipsis + (n-1)}}} is equal to {{{(n-1)*n/2}}}. 

It is the sum of a special arithmetic progression which is the sequence of the first (n-1) natural numbers. 
If you don't know it, see the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A> in this site. 

By combining the third addends of these trinomials, you will get {{{d^2*(0 + 1^2 + 2^2 + 3^2 + ellipsis + (n-1)^2)}}}.

This sum of squares of the first (n-1) natural numbers {{{1^2 + 2^2 + 3^2 + ellipsis + (n-1)^2}}} is equal to {{{((n-1)*n*(2n-1))/6}}}.

For the proof see, for example, the lesson <A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Mathematical-induction-for-sequences-other-than-arithmetic-or-geometric.lesson>Mathematical induction for sequences other than arithmetic or geometric</A> in this site. 

Now we can finalize our calculations.

{{{S[n]}}} = {{{a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+ ellipsis +(a+(n-1)d)^2}}} = 

= {{{n*a^2}}} + {{{2d}}}.{{{(n-1)*n)/2}}} + {{{d^2*((n-1)*n*(2n-1)/6)}}} = 

= {{{n*a^2}}} + {{{d*(n-1)*n}}} + {{{d^2}}}.{{{((n-1)*n*(2n-1))/6)}}}.
</pre>