SOLUTION: 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

Algebra ->  Sequences-and-series -> SOLUTION: 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      Log On


   

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
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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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
---------------------------------------------------- 
Let us consider one typical term %28a%2Bkd%29%5E2. It is
a%5E2 + 2%2Ak%2Ad + k%5E2%2Ad%5E2.

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%2Aa%5E2, right?

By combining the second addends of these trinomials, you will get 2d%2A%280+%2B+1+%2B+2+%2B+3+%2B+ellipsis+%2B+%28n-1%29%29.

If you know it, this sum 1+%2B+2+%2B+3+%2B+ellipsis+%2B+%28n-1%29 is equal to %28n-1%29%2An%2F2. 

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 d%5E2%2A%280+%2B+1%5E2+%2B+2%5E2+%2B+3%5E2+%2B+ellipsis+%2B+%28n-1%29%5E2%29.

This sum of squares of the first (n-1) natural numbers 1%5E2+%2B+2%5E2+%2B+3%5E2+%2B+ellipsis+%2B+%28n-1%29%5E2 is equal to %28%28n-1%29%2An%2A%282n-1%29%29%2F6.

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.

S%5Bn%5D =  = 

= n%2Aa%5E2 + 2d.%28n-1%29%2An%29%2F2 + d%5E2%2A%28%28n-1%29%2An%2A%282n-1%29%2F6%29 = 

= n%2Aa%5E2 + d%2A%28n-1%29%2An + d%5E2.%28%28n-1%29%2An%2A%282n-1%29%29%2F6%29.