Question

O valor de K, que torna a igualdade verdadeira é igual a:

Answer 1

Explicação passo-a-passo:

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Achando o valor de k na igualdade:

$\displaystyle \sum\limits_{i=1}^{5}(1+i)^{2}=35k+\displaystyle \sum\limits_{i=1}^{5}i^{2} \\\\\\(1+1)^{2} +(1+2)^{2}+(1+3)^{2}+(1+4)^{2}+(1+5)^{2}=35k+\displaystyle \sum\limits_{i=1}^{5}i^{2}\\\\\\2^{2}+ 3^{2}+4^{2}+5^{2}+6^{2}=35k+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}\\\\6^{2}=35k+1^{2} \\\\35k+1=36\\\\35k=36-1\\\\35k=35\\\\k=\dfrac{35}{35} \\\\\boxed{\boxed{k=1}}$

Resposta B

Answer 2

O valor de k que torna a igualdade verdadeira é k = 1. Item correto: B).

$\large\begin{array}{l}\!\displaystyle\sf\sum_{i\,=\:\!1}^{5}(1+i)^2=35^{}k+\!\:\!\sum_{i\,=\:\!1}^{5}i^2\\\\ \displaystyle\sf\sum_{i\,=\:\!1}^{5}(1+i)^2-\sum_{i\,=\:\!1}^{5}i^2=35k\\\\ \displaystyle\sf\sum_{i\,=\:\!1}^{5}\big[(1+i)^2-i^2\big]=35k\\\\\displaystyle\sf\sum_{i\,=\:\!1}^{5}\big[1+2i+i^2-i^2\big]=35k\\\\\displaystyle\sf\sum_{i\,=\:\!1}^{5}(1+2i)=35k\\\\\displaystyle\sf\sum_{i\,=\:\!1}^{5}1+\sum_{i\,=\:\!1}^{5}(2^{}i)=35k\\\\\sf 1\cdot5+2\cdot\!\displaystyle\sf\sum_{i\,=\:\!1}^{5}i=35k\end{array}$

Relembrando que

$\large\displaystyle\sf\sum_{i\,=\:\!1}^{5}i=1+2+3+4+5=15$

, chegaremos a:

$\large\begin{array}{l}\!\sf 1\cdot 5+2\cdot\!\displaystyle\sf\underbrace{\sf \sum_{i\,=\:\!1}^{5}i}_{15}=35k\\\\ \sf\!1\cdot 5+2\cdot 15=35k\\\\ \sf\!5+30=35k\\\\ \sf\!35k=35\\\\ \sf k=\dfrac{35}{35}\\\\ \!\boxed{\begin{array}{l}\sf k=1\end{array}}\end{array}$