sumatorias

UNIVERSIDAD NACIONAL DE HUANCAVELICAFACULTAD DE CIENCIAS DE INGENIERÍA

ESCUELA ACADÉMICO PROFESIONAL DE HUANCAVELICA

SUMATORIAS

1¿

Sabiendo que:

∑i=1

n

( f ( i )−f ( i−1 ) )=f (n )−f (0 )…1ra regla telesc ó pica

Sea : f ( i )=( i+1 )6

∑i=1

n

i5= [ ( i+1 )6−(i+1−1 )6 ]=(n+1 )6−1

∑i=1

n

i5=[ (i6+6 i5+15 i4+20 i3+15 i2+6 i+1 )−(i )6 ]= (n+1 )6−1

6∑i=1

n

i5+15∑i=1

n

i4+20∑i=1

n

i3+15∑i=1

n

i2+6∑i=1

n

i+∑i=1

n

1=(n+1 )6−1

6∑i=1

n

i5+15[ 6n5+15n4+10n3−n30 ]+20[ n4+2n3+n24 ]+15 [ 2n3+3n2+n6 ]+6 [ n2+n2 ]+ [n ]=(n+1 )6−1

6∑i=1

n

i5+[ 6n5+25n4+40n3+31n2+12n2 ]=[ (n6+6n5+15n4+20n3+15n2+6n+1 ) ]−1

ANÁLISIS MATEMÁTICO III Mag. Mat. César Castañeda Campos



6∑i=1

n

i5=[ (n6+6n5+15n4+20n3+15n2+6n+1 ) ]– 1−[ 6n5+25n4+40n3+31n2+12n2 ]6∑

i=1

n

i5=[ 2n6+12n5+30n4+40n3+30n2+12n−6n5−25n4−40n3−31n2+12n2 ]∑i=1

n

i5=[ 2n6+6 n5+5n4−n2

12 ]∑i=1

n

i5=[ n2(2n4+6n3+5n2−1)12 ]∑i=1

n

i5=[ n2(n+1)2(2n2+2n−1)12 ]2¿

Sabiendo que:

∑i=1

n


Sea : f ( i )=( i+1 )11




3¿

Aplicandola 1° regla telesc ó pica generalizada:




5¿∑i=1

n

ln2(i)

por propiedad de logaritmo tenemos :

∑i=1

n

ln2(i)=∑i=1

n

2 ln(i)

⇒∑i=1

n

2 ln(i)=2∑i=1

n

ln(i)

∑i=1

n

ln (i)=ln (1)+ln (2)+ ln(3)+ln(4)…… ..+ ln(n)

Por propiedad de logaritmo tenemos:




∑i=1

n

ln ( i)=ln(1.2 .3 .4 .5 .6…….n)

Por formula.

∑i=1

n

ln (i)=ln(n!)

∑i=1

n

ln2(i)=2 ln (n !)

6¿∑i=1

n

(i6)

Sabiendo que:

∑i=1

n


Sea : f (i )=(i+1 )7

∑i=1

n

i6=[ ( i+1 )7−(i+1−1 )7 ]=(n+1 )7−1

∑i=1

n

i5=[ (i7+7 i6+21 i5+35 i4+35 i3+21 i2+7 i+1 )−( i )7 ]= (n+1 )7−1




7∑i=1

n

i6+21∑i=1

n

i5+35∑i=1

n

i4+35∑i=1

n

i3+21∑i=1

n

i2+7∑i=1

n

i+∑i=1

n

1=(n+1 )7−1

7∑i=1

n

i6+21[ 2n6+6n5+5n4−n2

12 ]+35[ 6 n5+15n4+10n3−n30 ]+35[ n4+2n3+n24 ]+21[ 2n3+3n2+n6 ]+7[ n2+n2 ]+ [n ]=(n+1 )7−1

7∑i=1

n

i6+[ 210n6+1050n5+2100n4+2170n3+1260n2+410n60 ]=[ (n7+7 n6+21n5+35n4+35n3+21n2+7n+1 ) ]−1

7∑i=1

n

i6=[ (n7+7n6+21n5+35n4+35n3+21n2+7 n+1 ) ] –1−[ 210n6+1050n5+2100n4+2170n3+1260n2+410n60 ]

7∑i=1

n

i6=60n7

60+ 420n

6

60+ 1260n

5

60+2100n

4

60+ 2100n

3

60+1260n

2

60+ 420n60

−210n6

60−1050n

5

60−2100n

4

60−2170n

3

60−1260n

2

60+ 410n60

∑i=1

n

i6=[ 60n7+210n6+210n5−70n3+10n420 ]ANÁLISIS MATEMÁTICO III Mag. Mat. César Castañeda Campos



∑i=1

n

i6=[ 6n7+21n6+21n5−7n3+n42 ]∑i=1

n

i6=[ n(6n6+21n5+21n4−7n2+1)42 ]∑i=1

n

i6=[ n(n+1)(6n5+15n4+6n3−6n2−n+1)42 ]

8¿





9¿∑i=1

n

( 4 i25i )HALLANDO LA FORMULA :

4 i2

5i−4 ( i−1 )2

5i−1=4 i

2

5i−4 ( i−1 )2

5i−1

AplicandoSumatorias :

∑i=1

n

( 4 i25i −4 ( i−1 )2

5 i−1 )=∑i=1

n4 i2

5i−∑

i=1

n 4 ( i−1 )2

5i−1

TRABAJANDO EN ELPRIMERMIENBRO




f ( i )= 4 i2

5i⟹ f (i−1 )=4 (i−1 )2

5i−1

APLICANDO PROPIEDAD07 :

∑i=1

n

[ f ( i )−f ( i−1 ) ]=f (n )−f (0)

∴ f (n )−f (0 )=4 n2

5n−0

4n2

5n=∑

i=1

n4 i2

5i−∑

i=1

n 4 (i−1 )2

5i−1

4n2

5n=∑

i=1

n4 i2

5i−∑

i=1

n 4 (i2−2 i+1 )5i−1

4n2

5n=∑

i=1

n4 i2

5i−4∑

i=1

n (i2−2 i+1 )5i−1

4n2

5n=∑

i=1

n4 i2

5i−4∑

i=1

ni2

5i−1+4∑

i=1

n2 i5i−1

−4∑i=1

n15i−1

4n2

5n=∑

i=1

n4 i2

5i−∑

i=1

n4 i2

5i5−1+4∑

i=1

n2 i5i5−1

−4∑i=1

n1

5i5−1

4n2

5n=∑

i=1

n4 i2

5i−5∑

i=1

n4 i2

5 i+40∑

i=1

ni5i

−20∑i=1

n15i

4n2

5n=−4∑

i=1

n4 i2

5 i+40∑

i=1

ni5i

−20∑i=1

n15 i




4∑i=1

n4 i2

5i=40∑

i=1

ni5i

−20∑i=1

n15i

−4 n2

5n…(1)

∑i=1

ni5i… ( I )

Hallandouna formula para ( I )

∑i=1

ni5i

i

5i−i−15 i−1

= i

5i−i−15i−1


∑i=0

n

( i5i−i−15 i−1 )=∑

i=0

ni5i

−∑i=0

ni−15 i−1


f ( i )= i

5i⟹ f ( i−1 )=i−1

5i−1


∑i=1

n

[ f ( i )−f ( i−1 ) ]=f (n )−f (0)

∴ f (n )− f (0 )= n

5n−0




n5n

=∑i=0

ni5 i

−∑i=0

ni−15i−1

n5n

=∑i=0

ni5 i

−∑i=0

ni5i−1

+∑i=0

n15i−1

n5n

=∑i=0

ni5 i

−5∑i=0

ni5i

+5∑i=0

n15i

n5n

=−4∑i=0

ni5i

+5∑i=0

n15 i

∑i=0

ni5i

=54∑i=0

n15i

− n4.5n

Reemplazandoen :

4∑i=1

n4 i2

5i=40∑

i=1

ni5i

−20∑i=1

n15i

−4 n2

5n

4∑i=1

n4 i2

5i=40( 54∑i=0

n15i

− n4.5n )−20∑i=1

n15i

− 4n2

5n

4∑i=1

n4 i2

5i=50∑

i=0

n15i

−10 n5n

−20∑i=1

n15i

−4 n2

5n

4∑i=1

n4 i2

5i=30∑

i=0

n15i

−10 n5n

−4n2

5n…(2)

∑i=0

n15i… ( II)



ESCUELA ACADÉMICO PROFESIONAL DE HUANCAVELICAHallandouna formula para ( II )

∑i=1

n15i

1

5i− 1

5i−1= i

5i− 1

5i−1


∑i=0

n

( 15i− 15i−1 )=∑i=0

n15 i

−∑i=0

n15i−1


f ( i )= 1

5i⟹ f ( i−1 )= 1

5 i−1


∑i=1

n

[ f ( i )−f ( i−1 ) ]=f (n )−f (0)

∴ f (n )− f (0 )= 1

5n−0

15n

=∑i=0

n15 i

−∑i=0

n15i−1




15n

=∑i=0

n15 i

−5∑i=0

n15i

∑i=0

n15i

= −14.5n

Reemplazandoen2 :

4∑i=1

n4 i2

5i=30∑

i=0

n15i

−10 n5n

−4n2

5n…(2)

4∑i=1

n4 i2

5i=30( −1

4.5n )−10 n5n−4n2

5n

4∑i=1

n4 i2

5i=(−304.5n )−10 n5n−4 n

2

5n

∑i=1

n4 i2

5i=−3016 ( 15n )−104 n

5n−n2

5n

∑i=1

n4 i2

5i=−15n (358 +n2)




10¿∑i=1

n

( sen (i ) )

Usaremos laidentidad : sen (a ) . sen (b )=12

[cos (a−b )−cos (a+b ) ]

Sea :a=i ;b=12

Aplicamos sumatoriasaambos t é rminos

∑i=1

n

(sen ( i ) . sen ( 12 ))=¿12∑i=1

n [cos (i−12 )−cos( i+ 12 )]¿sen( 12 )∑i=1

n

( sen (i ) )=¿−12∑i=1

n [cos(i+12 )−cos (i−12 )]¿−2 sen ( 12 )∑i=1

n

(sen ( i) )=¿∑i=1

n [cos (i+ 12 )−cos (i−12 )]¿

Resolvemos el segundomiembro de la ecuaci ón ,usando :




∑i=1

n


Sea : f ( i )=cos (i+12 ); f ( i−1 )=cos (i−12 )∑i=1

n [cos( i+ 12 )−(cos (i−12 ))]=cos(n+12 )−cos ( 12 )Por lo tanto :

∑i=1

n

( sen ( i ) )=¿−∑i=1

n [cos (i+ 12 )−cos (i−12 )]2 sen ( 12 )

¿

∑i=1

n

( sen ( i ) )=¿−cos (n+ 12 )−cos( 12 )

2 sen ( 12 )¿




11¿∑i=1

n

(i7 )

Sabiendo que:

∑i=1

n


Sea : f ( i )=( i+1 )8

∑i=1

n

( ( i+1 )8−i8 )=(n+1 )8−1

∑i=1

n

( i8+8i7+28 i6+56 i5+70 i4+56 i3+28 i2+8i+1−i8 )=n8+8n7+28n6+56n5+70n4+56n3+28n2+8n+1−1

∑i=1

n

(8 i7+28 i6+56 i5+70i4+56i3+28 i2+8 i+1 )=n8+8 n7+28n6+56n5+70n4+56n3+28n2+8n

8∑i=1

n

i7+28∑i=1

n

i6+56∑i=1

n

i5+70∑i=1

n

i4+56∑i=1

n

i3+28∑i=1

n

i2+8∑i=1

n

i+∑i=1

n

1=n8+8n7+28n6+56n5+70n4+56n3+28n2+8n

8∑i=1

n

i7+28( 6n7+21n6+21n5−7n3+n42 )+56 ( 2n6+6n5+5n4−n2

12 )+70 (6 n5+15n4+10n3−n30 )+56 ( n4+2n3+n24 )+28( 2n3+3n2+n6 )+8 ( n2+n2 )+n=n8+8n7+28n6+56n5+70n4+56n3+28n2+8n

8∑i=1

n

i7=n8+8n7+28n6+56n5+70n4+56n3+28n2+8n−n−4 (n2+n )−143

(2n3+3n2+n )−14 (n4+2n3+n2 )−73

(6n5+15n4+10n3−n )−143

(2n6+6n5+5n4−n2)−23

(6n7+21n6+21n5−7n3+n )




8∑i=1

n

i7=n8+8n7+28n6+56n5+70n4+56n3+28n2+8n−n−4 n2−4 n−283n3−42

3n2−14

3n−14n4−28n3−14n2−14 n5−35n4−70

3n3+ 7

3n−28

3n6−28n5−70

3n4+ 14

3n2−4n7−14n6−14n5+ 14

3n3−2

3n

8∑i=1

n

i7=n8+n7 (8−4 )+n6(28−283 −14 )+n5 (56−14−28−14 )+n4(70−14−35−703 )+n3(56−283 −28−703

+ 143 )+n2(28−4−423 −14+14

3 )+n (8−1−4−143 + 73−23 )

8∑i=1

n

i7=n8+4n7+ 143n6−7

3n4+ 2

3n2

8∑i=1

n

i7=3n8+12n7+14n6−7n4+2n2

3

∑i=1

n

i7=n2 (3n6+12n5+14 n4−7 n2+2 )

24

12¿



ESCUELA ACADÉMICO PROFESIONAL DE HUANCAVELICAAplicandola 1° regla telesc ó pica generalizada:




14¿

Aplicandola propiedad de logaritmo




15¿∑i=1

n

(cos (i ) )

Usaremos laidenti d ad : sen (a ) .cos (b ) ¿ 12

[sen (a−b )+sen (a+b ) ]

Sea :a=1 /2 ;b=i

Aplicamos sumatoriasaambos t é rminos

∑i=1

n [sen ( 12 ) .cos ( i)]=12∑i=1n [sen( 12−i)+sen (i+ 12 )]

sen( 12 )∑i=1n

(cos (i ) )=¿ 12∑i=1

n [ sen(i+ 12 )+sen(−(i−12 ))]¿sen( 12 )∑i=1

n

(cos (i ) )=¿ 12∑i=1

n [ sen(i+ 12 )−sen( i−12 )]¿2 sen ( 12 )∑i=1

n

(cos (i ) )=¿∑i=1

n [sen (i+12 )−sen(i−12 )]¿




Resolvemos el segundomiembro de la ecuaci ón ,usando :

∑i=1

n


Sea : f (i )=sen(i+ 12 ); f ( i−1 )=sen (i−12 )∑i=1

n [sen (i+12 )−sen(i−12 )]=sen (n+ 12 )−sen ( 12 )∑i=1

n [sen(i+12 )−sen(i−12 )]=sen(n+ 12 )−sen( 12 )Por lo tanto :

∑i=1

n

(cos (i ) )=¿−∑i=1

n [ sen(i+ 12 )−sen( i−12 )]2 sen ( 12 )

¿

∑i=1

n

(cos ( i ) )=¿sen (n+ 12 )−sen ( 12 )

2 sen( 12 )¿

16¿∑i=1

n

(i8 )




Sabiendo que:

∑i=1

n


Sea : f ( i )=( i+1 )9

∑i=1

n

( ( i+1 )9−i9 )=(n+1 )9−1

∑i=1

n

( i9+9i8+36 i7+84 i6+126 i5+126 i4+84 i3+36 i2+9i+1−i9 )=n9+9n8+36n7+84 n6+126n5+126n4+84n3+36n2+9n+1−1

∑i=1

n

(9 i8+36 i7+84 i6+126 i5+126 i4+84 i3+36 i2+9 i+1 )=n9+9n8+36n7+84 n6+126n5+126n4+84n3+36n2+9n

9∑i=1

n

i8+36∑i=1

n

i7+84∑i=1

n

i6+126∑i=1

n

i5+126∑i=1

n

i4+84∑i=1

n

i3+36∑i=1

n

i2+9∑i=1

n

i+∑i=1

n

1=n9+9n8+36n7+84n6+126n5+126n4+84 n3+36n2+9n




9∑i=1

n

i8+36( 3n8+12n7+14 n6−7n4+2n224 )+84 ( 6n7+21n6+21n5−7 n3+n42 )+126( 2n6+6 n5+5n4−n2

12 )+126( 6n5+15n4+10n3−n30 )+84 ( n4+2n3+n24 )+36( 2n3+3n2+n6 )+9( n2+n2 )+n=n9+9n8+36n7+84 n6+126n5+126n4+84n3+36n2+9n

9∑i=1

n

i8=n9+9n8+36n7+84n6+126n5+126 n4+84 n3+36n2+9n−n−92

(n2+n )−6 (2n3+3n2+n )−21 (n4+2n3+n2 )−215

(6n5+15n4+10n3−n )−212

(2n6+6n5+5n4−n2 )−2 (6 n7+21n6+21n5−7n3+n )−32

(3n8+12n7+14 n6−7 n4+2n2 )

9∑i=1

n

i8=n9+9n8+36n7+84n6+126n5+126 n4+84 n3+36n2+9n−n−92n2−9

2n−12n3−18n2−6n−21n4−42n3−21n2−126

5n5−63n4−42n3+ 21

5n−21n6−63n5−105

2n4+ 21

2n2−12n7−42n6−42n5+14 n3−2n−9

2n8−18n7−21n6+ 21

2n4−3n2

9∑i=1

n

i8=n9+n8(9−92 )+n7 (36−12−18 )+n6 (84−21−42−21 )+n5(126−1265 −63−42)+n4 (126−21−63−1052 + 212 )+n3 (84−12−42−42+14 )+n2(36−92−18−21+ 212 −3)+n(9−1−92−6+ 215 −2)

9∑i=1

n

i8=n9+ 92n8

+6 n7−215n5+2n3− 3

10n

9∑i=1

n

i8=10n9+45n8+60n7−42n5+20n3−3n

10

∑i=1

n

i8=n (10n8+45n7+60n6−42n4+20n2−3 )

90




17¿





19¿

Aplicandola 1° regla telesc ó pica :

21¿∑i=1

n

i9




Sabiendo que:

∑i=1

n


Sea : f ( i )=( i+1 )10




22¿

Aplicandola 1° regla telescópica generalizada:




24¿

Aplicandola 1° regla telescópica :




28¿∑i=1

n

senh (i )

RESOLUCION :

cos h ( i+1 )−cosh ( i−1 )=cos h ( i+1 )−cosh ( i−1 )…1

Por la identidad.

cosh (A+B )=cosh ( A ) cosh (B )+senh (A ) sen h (B )…2

cosh (A−B )=cosh ( A )cos h (B )−sen h (A ) senh (B )…3

Restando 2 y 3 se tiene.

cosh (A+B )−cosh ( A−B )=2 senh ( A ) senh (B )

Remplazando en 1 se tiene.

cos h ( i+1 )−cosh ( i−1 )=2 senh ( i ) senh (1 )

Aplicando sumatorias a ambos miembros se tiene.

∑i=1

n

[cosh (i+1 )−cosh ( i−1 ) ]=2 senh (1 )∑i=1

n

sen h (i )




Analizando el primer miembro.

f (i+1 )=cosh ( i+1 ) , f (i−1)=cosh ( i−1 )

Entonces aplicamos la segunda regle telescópica.

⟹ f (n+1 )=cosh (n+1 ) , f (n)=cosh (n ) , f (1)=cosh (1 ) , f (0)=1

∑i=1

n

[cosh (i+1 )−cosh ( i−1 ) ]=2 senh (1 )∑i=1

n

sen h (i )

cosh (n+1 )+cosh (n )−cosh (1 )−1=2 sen h (1 )∑i=1

n

senh ( i )

∑i=1

n

senh (i )= cosh(n+1 )+cosh (n )−cosh (1 )−1

2 senh (1 )




29¿∑i=1

n

cosh (i )

RESOLUCION

senh (i+1 )−sen h (i−1 )=senh ( i+1 )−sen h (i−1 )… .1

Por la identidad.

senh (A+B )=senh ( A ) cosh (B )+cos h ( A ) senh (B )……… ..2

senh (A−B )=senh ( A )cos h (B )−cosh ( A ) senh (B )………….3

Restando 2 y 3 se tiene.

senh (A+B )−senh ( A−B )=2cosh (A ) sen h (B )

Remplazando en 1 se tiene.

senh (i+1 )−sen h (i−1 )=2cosh ( i ) senh (1 )

Aplicando sumatorias a ambos miembros se tiene.

∑i=1

n

[senh (i+1 )−senh ( i−1 ) ]=2 senh (1 )∑i=1

n

cosh ( i )

Analizando el primer miembro.

f (i+1 )=senh ( i+1 ) , f (i−1)=senh ( i−1 )




Entonces aplicamos la segunda regle telescópica.

⟹ f (n+1 )=senh (n+1 ) , f (n )=senh (n ) , f (1)=senh (1 ) , f (0 )=0

∑i=1

n

[senh (i+1 )−senh ( i−1 ) ]=2 senh (1 )∑i=1

n

cosh ( i )

senh (n+1 )+senh (n )−senh (1 )=2 senh (1 )∑i=1

n

cosh ( i )

∑i=1

n

cosh ( i )= senh (n+1 )+senh (n )−senh (1 )2 senh (1 )


sumatorias

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