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

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

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

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

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

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

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

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

$$\sum_n = n + 3n^2(n+1) - n(n+1)(2n+1)$$

$$\sum_n = n + 3n^3 + 3n^2 - [2n^3 + 3n^2 + n]$$

$$\sum_n = (3-2)n^3 + (3-3)n^2 + (1-1)n = n^3 + 0 + 0$$

$$\sum_n = n^3$$

Therefore:

$$S_n = n^3$$