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Stefan......
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Pe: 2020-08-17T22:19:17+03:00 In: In:

va rog frumos …

Ma puteti ajuta ?

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  1. Best Answer
    Suport

    a)1-\frac12+\frac13-\frac14+...+\frac1{2009}-\frac1{2010}+\frac1{2011}=\frac{1}{1006}+\frac{1}{1007}+\frac1{1008}+...+\frac1{2011}
    Observam ca fractiile cu numitorul impar(\frac1{1007}, \frac1{1009}…, \frac1{2011}) din membrul drept se afla si in membrul stang, cu acelasi semn. Scadem aceste numere. Ramanem cu:
    1-\frac12+\frac13-\frac14+...+\frac1{1005}-\frac1{1006}-\frac{1}{1008}-...-\frac{1}{2010}=\frac{1}{1006}+\frac1{1008}+...+\frac1{2010}
    A 2-a parte a membrului stang contine numerele din membrul drept. Scadem tot membrul drept:
    1-\frac12+\frac13-\frac14+...+\frac1{1005}-2(\frac1{1006}+\frac{1}{1008}+...+\frac{1}{2010})=0
    Simplificam acum fractiile din paranteza:
    1-\frac12+\frac13-\frac14+...+\frac1{1005}-(\frac1{503}+\frac{1}{504}+...+\frac{1}{1005})=0
    Numerele cu numitori impari din paranteza se mai afla o data in suma, cu semnul plus. Ii putem elimina:
    1-\frac12+\frac13-\frac14+...+\frac1{501}-\frac1{502}-\frac1{504}-\frac1{506}-...-\frac1{1004}-(\frac{1}{504}+\frac{1}{506}+...+\frac1{1004})=0
    Termenii din paranteza se repeta in suma cu semnul minus(acelasi pe care l-ar avea daca am desface paranteza):
    1-\frac12+\frac13-\frac14+...+\frac1{501}-\frac1{502}-2(\frac1{504}+\frac1{506}+...+\frac1{1004})=0
    Simplificam termenii din paranteza:
    1-\frac12+\frac13-\frac14+...+\frac1{501}-\frac1{502}-(\frac1{252}+\frac1{253}+...+\frac1{502})=0
    Termenii cu numitorul impar din paranteza se mai afla o data in suma cu semnul plus. Ii putem deci elimina:
    1-\frac12+\frac13-\frac14+...+\frac{1}{251}-\frac1{252}-\frac{1}{254}-...\frac1{502}-(\frac1{252}+...+\frac1{502})=0
    Termenii din paranteza se afla in suma cu semnul minus:
    1-\frac12+\frac13-\frac14+...+\frac{1}{251}-2(\frac1{252}+...+\frac1{502})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac{1}{251}-(\frac1{126}+...+\frac1{251})=0
    Termenii cu numitor impar din paranteza se mai afla in suma cu semnul plus, deci ii eliminam:
    1-\frac12+\frac13-\frac14+...+\frac{1}{125}-\frac1{126}-...-\frac{1}{250}-(\frac1{126}+...+\frac1{250})=0
    Termenii din paranteza se mai afla in suma cu semnul minus:
    1-\frac12+\frac13-\frac14+...+\frac{1}{125}-2(\frac1{126}+...+\frac1{250})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac{1}{125}-(\frac1{63}+...+\frac1{125})=0
    Din nou, termenii din paranteza cu numitor impar se simplifica:
    1-\frac12+\frac13-\frac14+...+\frac{1}{61}-\frac{1}{62}-\frac{1}{64}-...-\frac{1}{124}-(\frac1{64}+...+\frac1{124})=0
    Termenii din paranteza se mai afla in suma cu semnul minus:
    1-\frac12+\frac13-\frac14+...+\frac{1}{61}-\frac{1}{62}-2(\frac1{64}+...+\frac1{124})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac{1}{61}-\frac{1}{62}-(\frac1{32}+...+\frac1{62})=0
    Termenii cu numitor impar din paranteza se simplifica:
    1-\frac12+\frac13-\frac14+...+\frac1{31}-\frac1{32}-\frac1{34}-...-\frac1{62}-(\frac1{32}+...+\frac1{62})=0
    Termenii din paranteza se repeta in suma cu semnul minus:
    1-\frac12+\frac13-\frac14+...+\frac1{31}-2(\frac1{32}+...+\frac1{62})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac1{31}-(\frac1{16}+...+\frac1{31})=0
    Termenii din paranteza cu numitorul impar se simplifica:
    1-\frac12+\frac13-\frac14+...+\frac1{15}-\frac1{16}-\frac1{18}-...-\frac1{30}-(\frac1{16}+...+\frac1{30})=0
    Termenii din paranteza se repeta in suma cu semn minus:
    1-\frac12+\frac13-\frac14+...+\frac1{15}-2(\frac1{16}+...+\frac1{30})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac1{15}-(\frac1{8}+...+\frac1{15})=0
    Termenii cu numitor impar in paranteza se simplifica:
    1-\frac12+\frac13-\frac14+...+\frac1{7}-\frac1{8}-\frac1{10}-..-\frac1{14}-(\frac1{8}+...+\frac1{14})=0
    Termenii din paranteza se repeta cu semnul minus:
    1-\frac12+\frac13-\frac14+...+\frac1{7}-2(\frac1{8}+...+\frac1{14})=0
    Simplificam paranteza:
    1-\frac12+\frac13-\frac14+...+\frac1{7}-(\frac1{4}+...+\frac1{7})=0
    Suma mai are foarte putini termeni, asa ca ii putem scrie pe toti:
    1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac14-\frac15-\frac16-\frac17=0
    Simplificam \frac15 si \frac17:
    1-\frac12+\frac13-\frac14-\frac16-\frac14-\frac16=0
    1-\frac12+\frac13-2(\frac14+\frac16)=0
    1-\frac12+\frac13-(\frac12+\frac13)=0
    1-\frac12-\frac12=0
    1-1=0
    0=0

    Am ajuns la o expresie adevarata, deci egalitatea de la care am pornit este adevarata. O sa ma gandesc maine daca exista o solutie fara atat de multe calcule.

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  2. user (0)

    Si punctul b daca vrei. 

    • 0
  3. Suport

    b)Observam ca suma din membrul drept al punctului b seamana cu suma din membrul stang de la a). Ne punem intrebarea daca putem sa gasim o relatie asemanatoare celei de la punctul a) pentru suma de la punctul b). Raspunsul este ca da, anume:
    1-\frac12+\frac13-\frac14+...+\frac1{1333}-\frac1{1334}+\frac1{1335}=\frac1{668}+\frac1{669}+...+\frac1{1335}
    Demonstratia este aceeasi ca cea de la punctul a). O vei scrie fara prea multe explicatii.

    1-\frac12+\frac13-\frac14+...+\frac1{1333}-\frac1{1334}+\frac1{1335}=\frac1{668}+\frac1{669}+...+\frac1{1335}
    1-\frac12+\frac13-\frac14+...+\frac1{667}-(\frac1{668}+\frac1{670}+...+\frac1{1334})+(\frac1{669}+\frac1{671}+...+\frac1{1335})=\frac1{668}+\frac1{669}+...+\frac1{1335}

    1-\frac12+\frac13-\frac14+...+\frac1{667}-2(\frac1{668}+\frac1{670}+...+\frac1{1334})=0

    1-\frac12+\frac13-\frac14+...+\frac1{667}-(\frac1{334}+\frac1{335}+...+\frac1{667})=0

    1-\frac12+\frac13-\frac14+...+\frac1{333}-(\frac1{334}+\frac1{336}+...+\frac1{666})+(\frac1{335}+\frac1{337}+...+\frac1{667})-(\frac1{334}+\frac1{335}+...+\frac1{667})=0

    1-\frac12+\frac13-\frac14+...+\frac1{333}-2(\frac1{334}+\frac1{336}+...+\frac1{666})=0

    1-\frac12+\frac13-\frac14+...+\frac1{333}-(\frac1{167}+...+\frac1{333})=0

    1-\frac12+\frac13-\frac14+...+\frac1{165}-\frac1{166}-(\frac1{168}+\frac1{170}+...+\frac1{332})+(\frac1{167}+\frac1{169}+...+\frac1{333})-(\frac1{167}+...+\frac1{333})=0

    1-\frac12+\frac13-\frac14+...+\frac1{165}-\frac1{166}-2(\frac1{168}+...+\frac1{332})=0

    1-\frac12+\frac13-\frac14+...+\frac1{165}-\frac1{166}-(\frac1{84}+...+\frac1{166})=0

    1-\frac12+\frac13-\frac14+..+\frac1{83}-(\frac1{84}+\frac1{86}+...+\frac1{166})+(\frac1{85}+\frac1{87}...+\frac1{165})-(\frac1{84}+...+\frac1{166})=0

    1-\frac12+\frac13-\frac14+..+\frac1{83}-2(\frac1{84}+\frac1{86}+...+\frac1{166})=0

    1-\frac12+\frac13-\frac14+...+\frac1{83}-(\frac1{42}+...+\frac1{83})=0

    1-\frac12+\frac13-\frac14+...+\frac1{41}-2(\frac1{42}+\frac1{44}+...+\frac1{82})=0

    1-\frac12+\frac13-\frac14+...+\frac1{41}-(\frac1{21}+\frac1{22}+...+\frac1{41})=0

    1-\frac12+\frac13-\frac14+...-\frac1{20}-2(\frac1{22}+\frac1{24}+...+\frac1{40})=0

    1-\frac12+\frac13-\frac14+...-\frac1{20}-(\frac1{11}+\frac1{12}+...+\frac1{20})=0

    1-\frac12+\frac13-\frac14+...-\frac1{10}-2(\frac1{12}+\frac1{14}+...+\frac1{20})=0

    1-\frac12+\frac13-\frac14+...-\frac1{10}-(\frac1{6}+\frac1{7}+...+\frac1{10})=0

    1-\frac12+\frac13-\frac14+\frac15-2(\frac16+\frac18+\frac1{10})=0

    1-\frac12+\frac13-\frac14+\frac15-(\frac13+\frac14+\frac1{5})=0

    1-\frac12-\frac14-\frac14=0
    0=0

    • 1
  4. user (0)

    Mulțumesc mult!

    • 0
  5. Suport

    Am demonstrat mai sus ca membrul drept este egal cu:
    \frac1{668}+\frac1{669}+...+\frac1{1335}

    Observam ca 668+1335=2003. In acest caz, grupam termenii sirului astfel:
    (\frac1{668}+\frac1{1335})+(\frac1{669}+\frac1{1334})+(\frac1{670}+\frac1{1333})+...

    Fiecare paranteza are forma:
    \frac1{668+k}+\frac1{1335-k}, cu k natural. Deoarece suma are 1335-667=668, care este un numar par, deci toate numerele pot fi grupate. Cum toate perechile sunt de 2 termeni si avem in total 668 de numere, inseamna ca vom avea 668:2=334 de perechi, deci ultima pereche va fi \frac1{668+333}+\frac1{1335-333}, deoarece am inceput de la k=0.

    Sa analizam „termenul general”:
    \frac1{668+k}+\frac1{1335-k}=\frac{1335-k+668+k}{(668+k)(1335-k)}=\frac{1335+668}{(668+k)(1335-k)}=\frac{2003}{(668+k)(1335-k)}

    Suma devine:
    \frac{2003}{668*1335}+\frac{2003}{669*1334}+...+\frac{2003}{1001*1002}=2003(\frac{1}{668*1335}+\frac{1}{669*1334}+...+\frac{1}{1001*1002})
    Paranteza este o suma de termeni rationali, deci este rationala. Rezulta ca exista p si q naturale, cu (p, q)=1 astfel incat suma este egala cu:
    2003\cdot\frac{p}{q}. Atunci avem:
    \frac{m}{n}=\frac{2003p}{q}, deci m=2003pk, cu k natural, dexi m este multiplu de 2003.

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