User:Dylanwhs/dylwhs 2

$c=\pi r^{2}\,\!$

$e^{i\theta }=\cos \theta +i\sin \theta \,\!$

The numbers which are subtracted from the Root form a series $S_{n}=1+7+19+37+.....$

If one looks at the sum, the number of terms relates to the cube root of the sum itself.

1:~ 1 2:~ 1 + 7 3:~ 1 + 7 + 19 4:~ 1 + 7 + 19 + 37 5:~ 1 + 7 + 19 + 37 + 61 . . n:~ 1 + 7 + 19 + 37 + 61 + ....... +

Notice also that these terms can be written as

n:~ 1 + (1+6) + (1+18) + (1+36) + (1+60) + .... +

  = 1 + (1 + 6) + (1 +6 + 12) + (1+6+12+18) + (1+6+12+18+24) + ....
  = 1 + (1 + A) + (1 + B) + (1 + C) + (1 + D) +

Where

A = 6 B = 6 + 12 C = 6 + 12 + 18 D = 6 + 12 + 18 + 24

1 occurs n times
6 = 6.1 occurs n-1 times

12 = 6.2 occurs n-2 times 18 = 6.3 occurs n-3 times 24 = 6.4 occurs n-4 times

The term which occurs 1 times only will be 6.(n-1)

Thus the sum becomes

$S_{n}=1+n+6[1.(n-1)+2.(n-2)+3(n-3)+.....+(n-1)(n-(n-1))]$

$=\sum {r=1}^{n}1+6[1n-1^{2}+2n-2^{2}+3n-3^{2}+......+n(n-1)-(n-1)^{2}]$

$=n+6[n(1+2+3+....+n-1)]-6[1^{2}+2^{2}+3^{3}+....+(n-1)^{2}]$ $\sum _{n}=n+6n\sum {r=1}^{n}-1r-6\sum {r=1}^{n}-1r^{2}$

$\sum _{n}=n+6n(\sum {r=1}^{n}r-n)-6(\sum {r=1}^{n}r^{2}-n^{2})$

$\sum _{n}=n+6\sum {r=1}^{n}r-6n^{2}-6\sum {r=1}^{n}r^{2}+6n^{2}$

$\sum _{n}=n+6\sum {r=1}^{n}r-6\sum {r=1}^{n}r^{2}$

$\sum _{n}=n+6{\frac {1}{2}}n(n+1)-6{\frac {1}{6}}n(n+1)(2n+1)$

$\sum _{n}=n+3n^{2}(n+1)-n(n+1)(2n+1)$

$\sum _{n}=n+3n^{3}+3n^{2}-[2n^{3}+3n^{2}+n]$

$\sum _{n}=(3-2)n^{3}+(3-3)n^{2}+(1-1)n=n^{3}+0+0$

$\sum _{n}=n^{3}$

Therefore:

$S_{n}=n^{3}$