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Denis
07-17-2011, 02:37 PM
All about sum of squares starting later than at 1; like:
25 + 36 + 49 + 64 = 174.

f^2 + (f+1)^2 + .... + (n-1)^2 + n^2 = S

What is n in terms of S anf f ?

Have fun!

Subhotosh Khan
07-17-2011, 04:58 PM
All about sum of squares starting later than at 1; like:
25 + 36 + 49 + 64 = 174.

f^2 + (f+1)^2 + .... + (n-1)^2 + n^2 = S

What is n in terms of S anf f ?

Have fun!

S \ = \ \frac{n(n+1)(2n+1)}{6} \ - \ \frac{f(f-1)(2f-1)}{6}

That's a horrible equation - now what.....

Denis
07-17-2011, 05:05 PM
S \ = \ \frac{n(n+1)(2n+1)}{6} \ - \ \frac{f(f-1)(2f-1)}{6}
That's a horrible equation - now what.....
n = ?

Denis
07-18-2011, 02:20 PM
S \ = \ \frac{n(n+1)(2n+1)}{6} \ - \ \frac{f(f-1)(2f-1)}{6}

Right, so:
n(n + 1)(2n + 1) = 6S - f(f - 1)(2f - 1) ; simplify:
2n^3 + 3n^2 + n - u = 0
where:
u = 6S - f(f - 1)(2f - 1)

Solve:
n = (12a^2 - 6a + 1) / (12a)
where:
a = {[SQRT(108u^2 - 1)] / [8 * 3^(3/2)] + u/4}^(1/3)

Go try it here: www.numberempire.com/equationsolver.php (http://www.numberempire.com/equationsolver.php)

I put this up cause I had a hard time believing it could be so ugly/complicated!!