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

$S n = 1 + n + 6[1.(n - 1) + 2.(n - 2) + 3(n - 3) + ..... + (n - 1)(n - (n - 1))]$

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

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

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

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

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

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

∑	= n + 3n²(n + 1) − n(n + 1)(2n + 1)
n

∑	= n + 3n³ + 3n² − [2n³ + 3n² + n]
n

∑	= (3 − 2)n³ + (3 − 3)n² + (1 − 1)n = n³ + 0 + 0
n

∑	= n³
n

Therefore:

$S n = n 3$