Home/ Intrebari/Q 82762
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

classius
Pe: 2013-09-11T07:04:38+03:00 In: In:

Combinatorica

Determina ti n stiind ca {{n}\choose{1}}+{{n}\choose{3}}+{{n}\choose{5}}+...=1024.

Am observat ca 1024=2^{10} iar 2^{10}={{10}\choose{0}}+{{10}\choose{1}}+...+{{10}\choose{10}}

Conform formulei: {{n}\choose{k}}={{n-1}\choose{k-1}}+{{n-1}\choose{k}}, am rescris:
{{10}\choose{0}}+{{10}\choose{1}}={{11}\choose{1}
.
.
.
{{10}\choose{8}}+{{10}\choose{9}}={{11}\choose{9}}
iar {{10}\choose{10}}={{11}\choose{11}}

Deci am ramas cu egalitatea:
{{n}\choose{1}}+{{n}\choose{3}}+{{n}\choose{5}}+...={{11}\choose{1}}+{{11}\choose{3}}+...+{{11}\choose{11}}

Desi e destul de evident ca n=11… cum dovedesc asta?

Multumesc mult!

Similare

1 raspuns

  1. gunty
    maestru (V)

    Foloseste formula

        \[ C_n^1 + C_n^3 + ... = 2^{n - 1} \]

    si egalitatea devine

        \[ 2^{n - 1} = 2^{10} \]

    , de unde n-1=10, deci n=11.

Raspunde

Raspunde