CPhill

 \( \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]\)

Note that

i¹  = i

i²  = -1

i³  =  i * i²  =  i * -1  =  - i

i⁴  =  i² * i²  =  -1 * - 1  =  1

So....the sum of these  = 0

And this pattern continues for each partial sum of

i¹⁺⁴ⁿ, i²⁺⁴ⁿ, i^{3 + 4n} and i^{4 + 4n}       where  n is an integer ≥ 0

So......we  will have

0  +  0  + 0  +  .......+  0  + 0  +  i⁹⁷ + i⁹⁸ + i⁹⁹  =

i^{1+ 4(24)}  +  i ^{2 + 4(24)}  + i ^{3 + 4(24)}   =

[    i    ]     +  [    -1   ]        +    [  - i  ]    =

-1

A quadratic will have only one root  when the discriminant, b^2  - 4ac  = 0

4x^2 + 16x - 9    ⇒   16^2   - 4(4)(-9)  > 0  

2x^2 + 80x + 400  ⇒    80^2  - 4(2 )( 400) > 0   

x^2 - 6x - 9     ⇒   (-6)^2  - 4(1)(-9)  > 0 

4x^2 - 12x + 9   ⇒   (-12)^2  - 4 (4)(9)  =  0

-x^2 + 14x + 49  ⇒  (14)^2  - 4(-1)(49)  > 0

So

4x^2  - 12x +  9  =  0  factors as

(2x - 3) (2x - 3)  = 0

(2x - 3)^2  =  0

And this is true when   x  =  3/2

No table appears.......  

See the pic of the situation, here :

Note that in triangle ABF.....that  AF  is a base  and DG is  a segment parallel to this base....but a segment in a triangle that is parallel to a base splits the sides of the triangle proportionally

Therefore  :   BD / DA  =  BG/ GF

But D  is the midpoint of AB...so.... BD  = DA

But....this also means that  BG  =  GF

And  DGC  forms another triangle with DG  as a base  and EF a segment parallel to this base

And for the same reason as above :

DE / EC  =  GF / FC

But  E  is the midpoint of  CD....thus....DE  = EC

But...this also means that GF =  FC

But....it was proved that  BG =  GF

So.....by transitivity ...... FC  also  equals BG

Therefore    BG  = GF  =  FC

1 / √2    +  3 / √8   +  6 / √32   

1 / √2  +  3 / [ 2 √2]  +  6 / [ 4 √2 ]

1 / √2    +  3 / [ 2 √2]  + 3 / [ 2 √2 ]

2 / [ 2 √2]  +  6 / [ 2 √2]  

8 /  [ 2 √2 ]  

4  / √2  

4 *√2  /  [ √ 2 * √ 2 ]

4 * √2  /  2  

(4/2) √2

2√2

( 3x^2  - 2x  + 2  ) -  ( x^2 + 5x - 5 )  =   a negative outside a parenthese changes the signs on the terms inside

3x^2  - 2x   +  2   -  x^2  - 5x  +  5  =  combine like terms

3x^2 - x^2 - 2x - 5x  + 2 + 5  = 

2x^2  - 7x  + 7

( 4x  - 3  + 2x^2 )  +   (2x + 1)  =

4x + 2x  +  2x^2  -  3   +  1  =

2x^2 +  6x  -  2

 −6x³y² + 5x⁴ − 7z

The degree  will be the greatest sum of the exponents on any term

So....    3  + 2    =    degree 5

30 = 10x + 6y - 4x      simplify

30  = 6x + 6y       divide through by 6

5  = x  +  y

So....  b and d   are the answers  where the sum of the coordinates  = 5

Quartic =  highest power on x  = 4

"Tri" nomial  =  3 terms

So

Quartic Trinomial