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tiredstudent
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Pe: 2021-04-21T16:14:47+03:00 In: In:

ma puteti ajuta va rog la O2. …

ma puteti ajuta va rog la O2. Multumesc!

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  1. Best Answer
    maestru (V)

    \sum_{k=1}^n(aC_k^1+bC_{k+1}^2)=C_{n+1}^2\cdot\frac{4n+11}{6}
    a\sum_{k=1}^nC_k^1+b\sum_{k=1}^nC_{k+1}^2=\frac{(n+1)!}{(n-1)!\cdot2!}\cdot\frac{4n+11}{6}
    a\sum_{k=1}^n\frac{k!}{(k-1)!\cdot1!}+b\sum_{k=1}^n\frac{(k+1)!}{(k-1)!\cdot2!}=\frac{n(n+1)}{2}\cdot\frac{4n+11}{6}
    a\sum_{k=1}^nk+b\sum_{k=1}^nk(k+1)=\frac{n(n+1)(4n+11)}{12}
    Pentru prima suma avem o formula, formula lui Gauss anume faptul ca \sum_{k=1}^nk=\frac{n(n+1)}2. Aceasta formula este foarte cunoscuta si foarte usor de demonstrat.

    A 2-a suma o tratam astfel:
    \sum_{k=1}^nk(k+1)=\sum_{k=1}^n(k^2+k)=\sum_{k=1}^nk^2+\sum_{k=1}^nk

    Suma de k o stim(am mentionat formula mai sus). Exista o formula si pentru suma de patrate, dar eu n-am reusit niciodata sa o tin minte asa ca ii vom da o demonstratie. In tema ta poti sa scrii direct formula finala.

    Observam ca (k+1)^3-k^3=(k+1)(k^2+2k+1)-k^3=k^3+2k^2+k+k^2+2k+1-k^3=3k^2+3k+1. Atunci:
    \sum_{k=1}^n((k+1)^3-k^3)=\sum_{k=1}^n(3k^2+3k+1)
    Scriem dezvoltat suma din partea stanga:
    (n+1)^3-n^3+n^3-(n-1)^3+...+2^3-1^3=3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n1
    Observam ca toti termenii se anuleaza mai putin primul si ultimul:
    (n+1)^3-1=3\sum_{k=1}^nk^2+3\frac{n(n+1)}{2}+n
    n^3+3n^2+3n=3\sum_{k=1}^nk^2+\frac32(n^2+n)+n
    n^3+(3-\frac32)n^2+(3-\frac32-1)n=3\sum_{k=1}^nk^2
    n^3+\frac32n^2+\frac12n=3\sum_{k=1}^nk^2
    Impartind prin 3:
    \frac{n^3}3+\frac{n^2}{2}+\frac{n}6=\sum_{k=1}^nk^2
    Rescriem partea stanga astfel:
    \frac{2n^3+3n^2+n}{6}=\frac{n(2n^2+2n+1)}{6}=\frac{n(2n^2+n+n+1)}{6}=\frac{n(n(2n+1)+(n+1)}{6}=\frac{n(n+1)(2n+1)}{6}

    Ne intoarcem la egalitate:
    a\sum_{k=1}^nk+b\sum_{k=1}^nk+b\sum_{k=1}^nk^2=\frac{n(n+1)(4n+11)}{12}
    Inlocuim sumele cu formulele lor:
    (a+b)\frac{n(n+1)}{2}+b\frac{n(n+1)(2n+1)}6=\frac{n(n+1)(4n+11)}{12}
    Deoarece n este numar natural nenul(nu poate fi 0 deoarece  atunci suma de la 1 la 0 nu are avea sens), rezulta ca n(n+1) este diferit de 0 si deci putem imparti prin el:
    \frac{a+b}{2}+\frac{b}{6}(2n+1)=\frac{4n+11}{12}
    Inmultim cu 12:
    6(a+b)+2b(2n+1)=4n+11
    6a+6b+4bn+2b=4n+11

    2 polinoame sunt egale daca si numai daca coeficientii lor sunt egali, deci 6a+6b+2b=11 si 4b=4. Din a 2-a ecuatie obtinem ca b=1 si inlocuind in prima avem 6a+8=11, deci 6a=3 adica a=\frac12.

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