* Tính tổng:
B = $\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + $\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$
* Tính tổng: B = $\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + $\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$
By Eden
By Eden
* Tính tổng:
B = $\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + $\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$
⇒$\frac{1}{2}$[$\frac{2}{1.2.3}$+$\frac{2}{2.3.4}$+…+$\frac{2}{n(x+1)(n+2)}$]
⇒ B=$\frac{1}{2}$[$\frac{1}{1.2}$-$\frac{1}{2.3}$+$\frac{1}{2.3}$-…+$\frac{1}{n(n+1)}$-$\frac{1}{(n+1)(n+2)}$] ⇒B
=$\frac{1}{2}$[$\frac{1}{2}$-$\frac{1}{(x+1)(n+2)}$]
(Học tốt nhé!)
Đáp án:
`⇒B=\frac{1}{2}.(\frac{1}{2}-\frac{1}{(n+1)(n+2)})`
Giải thích các bước giải:
`B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+….+\frac{1}{n(n+1)(n+2)}`
`⇒B=\frac{1}{2}.(\frac{2}{1.2.3}+\frac{2}{2.3.4}+….+\frac{2}{n(n+1)(n+2)})`
`⇒B=\frac{1}{2}.(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+….+\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})`
`⇒B=\frac{1}{2}.(\frac{1}{2}-\frac{1}{(n+1)(n+2)})`