Alan

Do you mean find the consecutive integers which span zero (i.e. one negative and the other positive)?

If so, the following table should help:

Hmmm!   

n        3^(n-1)(n-1)                n3^n

0        3^(-1)(-1) = -1/3         0*3^0 = 0

1        3^(1-1)(1-1) = 0         1*3^1 = 3

2        3^(2-1)(2-1) = 3         2*3^2 = 18

...etc.

I get these to be:

s_n = n3ⁿ   and t_n = (5 - 2n)3ⁿ  where n starts from 0.  (i.e. n = 0, 1, 2, 3, ...)

Thanks Chris!

In general:  \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)  so we can also do it like this:

\((2x-\frac{3}{4}y^2)^3\rightarrow (2x)^3-3(2x)^2(\frac{3}{4}y^2)+3(2x)(\frac{3}{4}y^2)^2-(\frac{3}{4}y^2)^3\\ \rightarrow 8x^3-9x^2y^2+\frac{27}{8}xy^4-\frac{27}{64}y^6\)

where \(a=2x \text{ and } b=\frac{3}{4}y^2\)

.

!!!!!!!!!

As follows:

1/n + 2/n² = 0.28

Let x = 1/n

Then 2x² + x - 0.28 = 0  which factors as (2x + 1.4)(x - 0.2)

The negative solution is obtained from 2x + 1.4 = 0 or x = - 0.7

hence n = 1/x = -1/0.7  or  n = -10/7 or n ≈ -1.429

ABP = CBD  and EDQ = CDB so, because CBD = CDB we must have ABP = EDQ 

Hence 5x - 7 = 2x + 11  

           3x = 18

             x = 6°

Therefore ABP = 5*6 - 7 or ABP = 23°

ABP + ABD = 180°  so  ABD = 180 - 23  or  ABD = 157°

Like so: