Question

$\begin{align*} & \sum _{k=1}^{n}\left( 8-x_{k}\right) =8n-5\\ & \sum _{k=1}^{n}\left( \begin{matrix} a-y_{k}\end{matrix} \right) .x_{k}=0\end{align*}$

olduğuna göre,$\sum _{k=1}^{n}x_{k}.y_k$ ifadesinin eşiti ?

@cevap:5a

Answer 1

$$\sum_{k=1}^n(8-x_k)=8n-5\Rightarrow 8n-\sum_{k=1}^nx_k=8n-5\Rightarrow \sum_{k=1}^n x_k=5$$ olur. Benzer yolla $$\sum_{k=1}^n(a-y_k).x_k=0\Rightarrow \sum_{k=1}^n(ax_k-y_k.x_k)=0 \Rightarrow a\sum_{k=1}^nx_k-\sum_{k=1}^ny_k.x_k=0$$

$$5a-\sum_{k=1}^ny_k.x_k=0\Rightarrow \sum_{k=1}^ny_k.x_k=5a$$ olur.

Comments

eyvallah hocam eksik olmayın :))

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Önemli değil. Kolay gelsin.

Answer 2

$\displaystyle\sum_{k=1}^{n}(8-x_k)=8n-\displaystyle\sum_{k=1}^{n}x_k=8n-5$ imiş

$\displaystyle\sum_{k=1}^{n}x_k=5$ dir ozaman...

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$\displaystyle\sum_{k=1}^{n}(a-y_k).x_y=0$

$\displaystyle\sum_{k=1}^{n}a.x_k=\displaystyle\sum_{k=1}^{n}x_k.y_k$

$\displaystyle\sum_{k=1}^{n}x_k=5$ di ozaman 

bizden ıstenen ıfade 

$\displaystyle\sum_{k=1}^{n}[x_k.y_k]=5a$ imiş

Comments

eyvallah atom :)