Tính hợp lý
a) A = \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2016.2017}\right)\)
Cho P=\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{4033}{\left(2016.2017\right)^2}\)
Chứng minh rằng P<1
P=3 /1.22 +1/22.32+...+4033/20162.20172
P=1/1 -1/22 +1/22 -1/52 +...+1/20162 - 1/20172
P=1-1/20172 <1
vậy p<1
Tính A = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)+\left(-2-4-6-...-100\right)+\)\(\left(-1.2-2.3-3.4-...-99.100\right)\)
Tính
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
B=\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{5\cdot6}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}\)
\(A=\frac{5}{6}\)
\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(B=\frac{100}{2}\)
\(B=\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{99}\right)\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}\)
\(B=\frac{3\cdot4\cdot...\cdot100}{2\cdot3\cdot...\cdot99}\)
\(B=\frac{100}{2}\)
\(B=50\)
\(\left(1-\frac{1}{1.2}\right).\left(1-\frac{1}{2.3}\right)....\left(1-\frac{1}{2009.2010}\right)\)Tính
A = \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)
1tinh \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+....+\frac{89}{\left(44.45\right)^2}\)
2 tính \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).......\left(1-\frac{1}{12^2}\right)\)
1,tổng quát: (2k+1)/[k(k+1)^2]
=(2k+1)/k^2(k+1)^2=[(k+1)^^2-k^2]/k^2(k+1)^2=1/k^2-1/(k+1)^2
áp dụng vào ,kết quả =2024/2025
Hoàng Phúc bạn có thể giải chi tiết hơn một chút đc ko???
cho P= \(\frac{3}{\left(1.2\right)^2}\) +\(\frac{5}{\left(2.3\right)^2}\) +\(\frac{7}{\left(3.4\right)^2}\) +...+ \(\frac{4033}{\left(2016.2017\right)^2}\)
chứng minh P<1
\(P=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+.....+\frac{4033}{\left(2016.2017\right)^2}\)
\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+.....+\frac{2017^2-2016^2}{2016^2.2017^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{2016^2}-\frac{1}{2017^2}\)
\(=1-\frac{1}{2017^2}< 11\) (đpcm)
Bài này trong đề thi học kì 2 môn Toán lớp 6 trường Amsterdam năm 2016-2017 này. Mình 10 luôn hehe
\(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{29.30}\right)\)
GẤP ... GẤP ... GẤP CÁC BẠN
P = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{4003}{\left(2016.2017\right)^3}\)
Chứng minh rằng : P < 1
A = \(\frac{2018^{100}+2018^{96}+...+2018^4+1}{2018^{102}+2018^{100}+...+2018^2+1}\)
Chứng minh rằng : 4A < \(10111^6\)