.
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
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Let us consider one typical term . It is
+ + .
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 , right?
By combining the second addends of these trinomials, you will get .
If you know it, this sum is equal to .
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 Arithmetic progressions in this site.
By combining the third addends of these trinomials, you will get .
This sum of squares of the first (n-1) natural numbers is equal to .
For the proof see, for example, the lesson Mathematical induction for sequences other than arithmetic or geometric in this site.
Now we can finalize our calculations.
= =
= + . + =
= + + ..