Question

chứng tỏ rằng

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}>\dfrac{1}{4}\)

Answer 1

* Chứng tỏ

Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)

= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)

= \(\dfrac{4849}{19800}\)

Answer 2

* So sánh

\(\dfrac{4950}{19800}\) và \(\dfrac{1}{4}\)

\(\dfrac{1}{4}=\dfrac{4950}{19800}\)

Vì \(\dfrac{4950}{19800}=\dfrac{4950}{19800}\)

=> Tổng trên bằng với\(\dfrac{1}{4}\)

Answer 3

mình ko chắc là đúng

Answer 4

Giải:

(Đề sai rùi, phải là \(< \dfrac{1}{4}\) mới đúng, cho nên mik làm đề đúng).

Đặt tên cho dãy phân số trên là A, ta có:

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}.\)

\(2A=2\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right).\)

\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}.\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}.\)\(2A=\dfrac{1}{1.2}+\left(\dfrac{1}{2.3}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{3.4}-\dfrac{1}{3.4}\right)+\left(\dfrac{1}{4.5}-\dfrac{1}{4.5}\right)+...+\left(\dfrac{1}{98}-\dfrac{1}{98}\right)-\dfrac{1}{99.100}.\)\(2A=\dfrac{1}{1.2}+0+0+...+0-\dfrac{1}{99.100}.\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{99.100}.\)

\(2A=\dfrac{1}{2}-\dfrac{1}{9900}.\)

\(2A=\dfrac{4950}{9900}-\dfrac{1}{9900}.\)

\(2A=\dfrac{4949}{9900}.\)

\(\Rightarrow A=\dfrac{4949}{9900}:2.\)

\(\Rightarrow A=\dfrac{4949}{19800}.\)

- Quy đồng mẫu số của phân số \(\dfrac{1}{4}\) thành phân số có mẫu là 19800: \(\dfrac{1}{4}=\dfrac{1.4950}{4.4950}=\dfrac{4950}{19800}.\)

Vì: \(A=\dfrac{4949}{19800}< \dfrac{1}{4}=\dfrac{4950}{19800}.\)

\(\Rightarrow A< \dfrac{1}{4}.\)

Vậy ta thu được \(đpcm.\)

~ Học tốt!!! ~