Đáp án: $Q=\dfrac{8975}{5304}$
Giải thích các bước giải:
a.Dạng tổng quát số hạng của $Q$ là:
$$\dfrac{3n+1}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}$$
b.Ta có:
$\dfrac{3n+1}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}$
$=\dfrac{3n}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}+\dfrac{1}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
$=3\cdot \dfrac{1}{\left(n+1\right)\cdot \left(n+2\right)}+\dfrac12\cdot \dfrac{2}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
$=3\cdot \dfrac{\left(n+2\right)-\left(n+1\right)}{\left(n+1\right)\cdot \left(n+2\right)}+\dfrac12\cdot\dfrac{\left(n+2\right)-n}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
$=3\left(\dfrac1{n+1}-\dfrac1{n+2}\right)+\dfrac12\left(\dfrac1{n\left(n+1\right)}-\dfrac1{\left(n+1\right)\left(n+2\right)}\right)$
Ta có:
$Q=\dfrac4{1\cdot 2\cdot 3}+\dfrac7{2\cdot 3\cdot 4}+...+\dfrac{151}{50\cdot 51\cdot 52}$
$\to Q=\left(3\left(\dfrac1{1+1}-\dfrac1{1+2}\right)+\dfrac12\left(\dfrac1{1\left(1+1\right)}-\dfrac1{\left(1+1\right)\left(1+2\right)}\right)\right)+...+\left(3\left(\dfrac1{50+1}-\dfrac1{50+2}\right)+\dfrac12\left(\dfrac1{50\left(50+1\right)}-\dfrac1{\left(50+1\right)\left(50+2\right)}\right)\right)$
$\to Q=\left(3\left(\dfrac1{2}-\dfrac1{3}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}\right)\right)+...+\left(3\left(\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)\right)$
$\to Q=3\left(\dfrac1{2}-\dfrac1{3}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}\right)+...+3\left(\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)$
$\to Q=3\left(\dfrac12-\dfrac13+...+\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}+....+\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)$
$\to Q=3\cdot \left(\dfrac12-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{51\cdot 52}\right)$
$\to Q=\dfrac{8975}{5304}$
