Ensino Superior ⇒ Teorema das Linhas Tópico resolvido

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Usando o Teorema das Linhas calcule a seguinte soma. Justifique

[tex3]S=\sum_{k=1}^{20}(k+1)C_{20}^{k}[/tex3]

[MateusQqMDMOD]

Olá, machadolfm

[tex3]S=\sum_{k=1}^{20}(k+1)C_{20}^{k}[/tex3]

[tex3]S=\sum_{k=1}^{20} k C_{20}^{k} + \sum_{k=1}^{20} C_{20}^{k} [/tex3]

[tex3]S=\sum_{k=1}^{20} k C_{20}^{k} + \sum_{k=0}^{20} C_{20}^{k} \, - C_{20}^0 [/tex3]

[tex3]S=\sum_{k=1}^{20} k \frac{20!}{k!(20-k)!} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S=\sum_{k=1}^{20} \frac{20!}{(k-1)!(20-k)!} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S=\sum_{k=1}^{20} 20 \frac{19!}{(k-1)!(20-k)!} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S= 20\sum_{k=1}^{20} \frac{19!}{(k-1)!(20-k)!} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S=20 \sum_{k=1}^{20} C_{19}^{k-1} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S=20 \cdot 2^{19} + 2^{20} \, - C_{20}^0 [/tex3]

[tex3]S=11 \cdot 2^{20} - 1 [/tex3]

[MateusQqMDMOD]

Machado, sabemos que o Teorema das Linhas nos dá que [tex3]C_n^0 + C_n^1 + C_n^2 \, + \, ... \, + \,C_n^n = \sum_{k=0}^nC_{n}^{k} = 2^n[/tex3]