Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(A=\frac{2^{2017}}{2^{2017}}+\frac{2^{2016}}{2^{2017}}+\frac{2^{2015}}{2^{2017}}+...+\frac{2^1}{2^{2017}}+\frac{1}{2^{2017}}\)
\(=\frac{1+2^1+2^2+...+2^{2016}+2^{2017}}{2^{2017}}\)
Đặt: B=\(1+2^1+2^2+...+2^{2017}\)
\(\Leftrightarrow2B=2^1+2^2+2^3+....+2^{2017}+2^{2018}\)
\(\Leftrightarrow2B-B=2^{2018}-1\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\Rightarrow A=\frac{B}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Mik chỉ biết làm phần a thôi
b/ Sử dụng quy tắc: \(\frac{a+c}{b+c}< \frac{a}{b}\) với \(\left\{{}\begin{matrix}a;b;c>0\\a>b\end{matrix}\right.\)
\(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)
\(\Rightarrow B>A\)
\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+......+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{20}.\left(1+2+...+20\right)\)
\(=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(3+1\right).3}{2}+...+\frac{1}{20}.\frac{\left(20+1\right).20}{2}\)
\(=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{20+1}{2}\)
\(=1+\frac{1}{2}.\left(1+2+1+3+...+20+1\right)\)
\(=1+\frac{1}{2}.\left[\left(1+1+...+1\right)+\left(1+2+3+...+20\right)\right]\)
\(=1+\frac{1}{2}.\left[20+\frac{\left(20+2\right).19}{2}\right]\)
\(=1+\frac{1}{2}.\left[20+\frac{22.19}{2}\right]\)
\(=1+\frac{1}{2}.\left[20+11.19\right]\)
\(=1+\frac{1}{2}.\left[20+209\right]\)
\(=1+\frac{1}{2}.229\)
\(=\frac{2}{2}+\frac{229}{2}\)
\(=\frac{231}{2}\)
Tham khảo nhé~
A = 1 + 1/2 . (1 + 2) + 1/3 . (1 + 2 + 3) + ... + 1/20 . (1 + 2 + 3 + ... + 20)
= 3/2 + 3/2 + 4/2 + ... + 21/2
= 2 + 3 + 4 + ... + 21/2
= 230/2
= 115/1 = 115