CÁLCULO
DE “n” EN FUNCIÓN DE “N”
( P / 100 )
* N = 1 * n + 2 * ( n – 1 ) + 3 * ( n – 2 ) +..... + ( n –
2 ) * 3 + ( n – 1 ) * 2 + n * 1
i
= n
=
∑ [ i . [ n – ( i – 1 ) ]
i
= 1
i
= n i = n
=
∑ ( i . n ) – ∑ [ i * ( i – 1 ) ]
i
= 1 i = 1
i
= n
=
[ ( 1 + n ) / 2 ] * n 2 – ∑ { i * ( i – 1 ) ]
i
= 2
i
= n – 1
=
[ ( 1 + n ) / 2 ] * n 2 – ∑ [ i * ( i + 1 ) ]
i
= 1
i = n – 1
=
n 2 * [ ( n + 1 ) / 2 ] – { [ 1 + ( n – 1 ) ] * (n –
1 ) / 2 } – ∑ i 2
i = 1
i
= n – 1
=
n 2 * [ ( n + 1 ) / 2 ] – [ ( n 2 – n ) / 2
] – ∑ i 2
i
= 1
i
= n – 1
=
( n 3 + n 2 – n 2 + n ) / 2 –
∑ i 2
i = 1
=
( n 3 + n ) / 2 – { [ n * ( n + 1 ) * ( 2 * n + 1 )] /
6 – n 2 }
=
{ 3 * n 3 + 3 * n + 6 * n 2 – ( n
2 + n ) * ( 2 * n + 1 ) } /6
=
( 3 * n 3 + 3 * n + 6 * n 2 – 2 * n 3
– 2 * n 2 – n 2 – n ) / 6
=
( n 3 + 3 * n 2 + 2 * n ) / 6
=
[ n * ( n 2 + 3 * n + 2 ) ] / 6
=
[ n * ( n + 1 ) * ( n + 2 ) ] / 6
i
= 2 i = n
= ∏ [
n + i ] = ∑ [ i . [ n – ( i – 1 ) ] = ( P / 100 ) * N
i = 0 i = 1
=
1 * n + 2 * ( n – 1 ) + 3 * ( n – 2 ) +..... + ( n – 2 ) *
3 + ( n – 1 ) * 2 + n * 1
=
(1, 2., 3 … (n-1), n) * (n, (n-1), (n-2), … 3, 2, 1)
T = M * N T
matricialmente,
donde
M = (1, 2., 3 … (n-1), n) Є
М 1 * n y
N = (n, (n-1), (n-2), … 3, 2, 1) Є
М n * 1
CONVERSIÓN
DE UNA SERIE CUADRÁTICA EN UNA SERIE ARITMÉTICA A(1,1) Y
VICEVERSA
OBSERVACIÓN:
i
= n i = n
∑ i
– ∑ i 2
i = 1
i = 1
=
[ ( 1 + n ) / 2 ] * n – n * ( n + 1 ) * ( 2 * n
+ 1 )] / 6
i = n
=
∑ i * [ 1– ( 2 * n + 1 )] / 3 ]
i
= 1
i = n
=
∑ i * [ 2 ( 1 – n )] / 3 ]
i = 1
i
= n i = n i = n
=
(2 / 3 ) * ( 1 – n ) * ∑ i = ∑ i – ∑
i 2
i
= 1 i = 1 i
= 1
De
aquí deducimos
i
= n i = n i = n
∑ i
2
= ∑ i – (2 / 3 ) * ( 1 – n ) * ∑ i
i
= 1 i = 1 i = 1
i
= n
=
∑ i * [ 1 – (2 / 3 ) * ( 1 – n ) ]
i
= 1
i
= n
=
∑ i * ( 1 + 2 * n ) / 3 →
i
= 1
i = n i = n
haciendo
A = ∑ i y C = ∑ i 2 con series aritmética
A=A(1, 1) y cuadrática respectivamente es
i
= 1 i = 1
i
= n
i = n i = n
C
= ( 1 / 3 ) * ( 1 + 2 * n ) * A
(serie aritmética A(1,1) y C = ∑ i 2 = ∑ [ i – (2
/ 3 ) * ( 1 – n ) * ∑ i ]
i
= 1 i = 1 i = 1
DEMOSTRACIÓN
DE A – C ≠
A / C
Lo
hago por reducción al absurdo.
De
C = ( 1 / 3 ) * ( 1 + 2 * n ) * A tenemos
a)
A – C = A
* [ 1 – ( 1 / 3 )
* ( 1 + 2 * n
) ] = [ 2 * ( 1 – n
) / 3 ] * A
=
[ 2 * ( 1 –
n
) / 3 ] * [ n * ( n + 1 ) / 2] = n * ( 1 –
n 2
) / 3 →
n
* ( – n
2
+ 1 )
A
– C =_______________
3
3
b)
A / C =
[ 1 / 3 * ( 1 + 2 * n)
] - 1 =
3 / ( 2 * n+ 1 ) = ________ →
2
* n + 1
A
/ C = 3 / ( 2 * n+ 1 )
Entonces
si A – C =
A / C →
*
A * [ 2 * ( 1 –
n
) / 3 ] = 3/ ( 2 * n + 1 ) →
A = [
3 / ( 2 * n + 1 ) ] /
( 1
/ 3 ) * [ 2 * ( 1 –
n
) ]
=
9 / [ ( 2 * n + 1 ) * [ 2 * ( 1 – n
) ] → A = 9 / [ ( 2 * n + 1 ) * ( –
2 * n + 2 ) ] → (*1 )
*
A = C * ( A – C ) → A = C * A – C 2
→ C 2
– C * A + A = 0 → C = [ A ±
√
( A 2 –
4 * A ) ] / 2
=
[ A ± √
A * √
( A – 4 ) ] / 2 → C = [ A ±
√
A * √
( A – 4 ) ] / 2 → (*2 )
De
(*1
) →
I
= n 9 9
A = ∑
i = 9 / [ ( 2 * n +
1) ( - 2 * n + 2 ) ] = ___________________= ____________________
i
= 1 2 * ( - 2 * n 2
+ n + 1 ) -
4 * ( n + ½ ) * ( n - 1 )
I = n 9
→ ∑
i =
9 / [ - 4 * ( n + ½ ) * ( n – 1 ) ] = __________________ no es aritmética
i = 1 -
4 * ( n + ½ ) * ( n - 1 )
De
(*2
)
i = n
C
= ∑ i 2 = [ A ±
√A
*√(
A – 4 ) ] / 2 →
3 casos:
i
= 1
i = n i = n i = n
a1)
C = [ A +
√A
*√(
A – 4 ) ] / 2 = [
∑ i
+
√
∑ i
*√(
∑ i
– 4 ) ] / 2 no
es cuadratica
i
= 1 i = 1 i = 1
i
= n i = n i = n
a2)
C = [ A –
√A
*√(A
– 4 ) ] / 2 = [ ∑
i +
√
∑ i
*√(
∑ i
– 4 ) ] / 2 no
es cuadratica
i = 1 i = 1 i = 1
b)
C = 1 / 3 * ( 1 + 2 * n )
* A = [ 1 / 3 * ( 1 + 2 * n ) ] * { 9 /
[ – 4 * ( n +1/2 ) * (n – 1) ] }
=
[ ( 1 + 2 * n ) * 9 ] / { 3 * [ ( – 4) * ( n +1/2 ) * (n – 1) ]
}
–
3 * ( 2 * n + 1 )
=
( – 3/ 4 ) * ( 2 * n + 1 ) / ( n +1/2 ) * (n – 1) = ___________________
4 *
( n +1/2 ) * (n –
1)
i
= n –
3 * ( 2 * n + 1 )
→ ∑
i 2
=
=
[
– 3 * ( 2 * n + 1 )
]
/ [
4 *
( n +1/2 ) * (n –
1) ] = ___________________
i
= 1 4 *
( n +1/2 ) * (n –
1)
no
es cuadratica
EXPONENCIALES Y
LOGARITMOS
i = n
∑ { i . [ n - ( i – 1
) ] }
i
= 1i = n i
. [ n - ( i – 1 ) ]
exp = ∏
exp
i
= 1
n + i
i = n
6
n
- ( n + 1 ) . ( n + 2 )
∏
i
= 1
6
C n
+ 2,, n - 1
=
exp = exp
= exp
i
= n
i = n
log ∏ {
i . [ n - ( i – 1 ) ] } = ∑
log {
i . [ n - ( i – 1
) ] }
i = 1
i
= 1
