Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=4/3+9/8+16/15+..............+4064256/4064255
A=1+1/3+1+1/8+1/15+...............+1/4064255
A=(1+1+...+1)+(1/3+1/8+...+1/406255) (có 2015 số 1)
A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)
A=2015+1/2(1+1/2-1/2016-1/2017)
A=2015,749504
k cho mình nhé mình k lại cho
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)
=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)
=\(\frac{50.2}{1.51}\)
=\(\frac{100}{51}\)
\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)
\(B=\frac{\left(2.3.4...150\right)\left(2.3.4...150\right)}{\left(1.2.3...149\right)\left(3.4.5...151\right)}\)
\(B=\frac{\left(1.2.3...149\right).150.2.\left(3.4.5...150\right)}{\left(1.2.3...149\right).\left(3.4.5...150\right).151}\)
\(B=\frac{300}{151}\)
Theo quy luật mà mình nhận thấy thì 20112 phải sửa thành 20122 bạn ạ!
Đặt \(A=\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2012^2}\)
\(\Leftrightarrow A=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+\frac{4^2+1}{4^2}+...+\frac{2012^2+1}{2012^2}\)
\(\Leftrightarrow A=1+\frac{1}{2^2}+1+\frac{1}{3^2}+1+\frac{1}{4^2}+...+1+\frac{1}{2012^2}\)
\(\Leftrightarrow A=\left(1+1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
\(\Leftrightarrow A=2011+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
Có: \(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2012}\)
\(\Rightarrow A=2011+B< 2011+1-\frac{1}{2012}\)
\(\Rightarrow A< 2012-\frac{1}{2012}< 2013\)
Ta có đpcm