rút gọn
A= \(\frac{3}{\left(1.2^2\right)}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{n\left(n+1\right)^2}\)
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
a) Phân tích đa thức thành nhân tử: x(x+2)(x2+2x+2)+1
b) Rút gọn biểu thức: A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{[n\left(n+1\right)]^2}\)
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\(x\left(x+2\right)\left(x^2+2x+2\right)+1\)
\(=\left(x^2+2x\right)\left(x^2+2x+2\right)+1\)
Đặt: \(x^2+2x=t\)
khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)
\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)
Khi đó:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Rút gọn C = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Rút gọn: \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+....+\frac{2n+1}{\left(n.\left(n+1\right)\right)^2}\)
Cho a1, a2, a3,........., a2016 là các STN và tổng chúng chia hết cho 3. CMR: A=a13+a23+..............+a20163 chia hết cho 3.
Tính giá trị \(C=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)^2\right]}\)
Mỗi lần bạn lên OLM là toàn đang những câu hỏi cực khó
em mới học lớp 6 thôi chưa đủ trình để làm bài này nha
xin lỗi, nếu biết em nhất đình sẽ trả lời
help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
1. Chứng minh : B = \(\left(1-\frac{2}{6}\right).\left(1-\frac{2}{12}\right).\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
2. cho M = \(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+\frac{1}{5.\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
N = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}\)
Rút gọn \(\frac{M}{N}\)
tính
A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(3.2\right)^2}+.....+\frac{2n+1}{\left[n.\left(n+1\right)\right]^2}\)
Ta có :
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\frac{4-1}{1^2.2^2}+\frac{9-4}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
Chúc bạn học tốt ~
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(3.2\right)^2}+...+\frac{2n+1}{\left[n.\left(n+1\right)\right]^2}\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{2n+1}{n^2.\left(n+1\right)^2}\)
\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)
\(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(A=1-\frac{1}{\left(n+1\right)^2}\)
mk chỉ làm được đến đấy thôi
Rút gọn biểu thức sau:
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\) (n nguyên dương)
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)