$\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + … + $\frac{1}{n.(n+1).(n+2)}$ Mn giúp e với ạ , tối e nộp e ạ 05/09/2021 Bởi Delilah $\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ + … + $\frac{1}{n.(n+1).(n+2)}$ Mn giúp e với ạ , tối e nộp e ạ
`1/{1.2.3}+1/{2.3.4}+…+1/{n(n+1)(n+2)}` `=1/2(2/{1.2.3}+2/{2.3.4}+…+2/{n(n+1)(n+2)})` `=1/2(1/1.2-1/2.3+1/2.3-1/3.4+…+1/{n(n+1)}-1/{(n+1)(n+2)})` `=1/2(1/1.2-1/{(n+1)(n+2)})` `=1/4-1/{2(n+1)(n+2)}` `={2(n+1)(n+2)-4}/{8(n+1)(n+2)}` Bình luận
Đáp án: Giải thích các bước giải: đề $2S= \dfrac{2}{1. 2. 3}+ \dfrac{2}{2. 3. 4}+ \dfrac{2}{3. 4. 5}+ …+ \dfrac{2}{n (n+ 1). (n+ 2)}$ $2S= \dfrac{1}{1. 2}- \dfrac{1}{2. 3}+ \dfrac{2}{1. 2. 3}- \dfrac{1}{3. 4}+ \dfrac{1}{3. 4}- \dfrac{1}{4. 5}+…+ \dfrac{1}{n.(n+ 1)}- \dfrac{1}{(n+ 1). (n+ 2)}$ $2S= \dfrac{1}{2}- \dfrac{1}{(n+ 1). (n+ 2)}$ $S= \dfrac{1}{4}- \dfrac{1}{(n+ 1). (n+ 2): 2}$ Bình luận
`1/{1.2.3}+1/{2.3.4}+…+1/{n(n+1)(n+2)}`
`=1/2(2/{1.2.3}+2/{2.3.4}+…+2/{n(n+1)(n+2)})`
`=1/2(1/1.2-1/2.3+1/2.3-1/3.4+…+1/{n(n+1)}-1/{(n+1)(n+2)})`
`=1/2(1/1.2-1/{(n+1)(n+2)})`
`=1/4-1/{2(n+1)(n+2)}`
`={2(n+1)(n+2)-4}/{8(n+1)(n+2)}`
Đáp án:
Giải thích các bước giải:
đề
$2S= \dfrac{2}{1. 2. 3}+ \dfrac{2}{2. 3. 4}+ \dfrac{2}{3. 4. 5}+ …+ \dfrac{2}{n (n+ 1). (n+ 2)}$
$2S= \dfrac{1}{1. 2}- \dfrac{1}{2. 3}+ \dfrac{2}{1. 2. 3}- \dfrac{1}{3. 4}+ \dfrac{1}{3. 4}- \dfrac{1}{4. 5}+…+ \dfrac{1}{n.(n+ 1)}- \dfrac{1}{(n+ 1). (n+ 2)}$
$2S= \dfrac{1}{2}- \dfrac{1}{(n+ 1). (n+ 2)}$
$S= \dfrac{1}{4}- \dfrac{1}{(n+ 1). (n+ 2): 2}$