Rút gọn :\(B=1^2+3^2+5^2+...+\left(2n+1\right)^2\)
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Rút gọn:
B= \(\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}...\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(=\frac{1^2}{\left(2-1\right)\left(2+1\right)}.\frac{3^2}{\left(4-1\right)\left(4+1\right)}...\frac{\left(2n+1\right)^2}{\left(2n+2-1\right)\left(2n+2+1\right)}\)
\(=\frac{1}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)
\(=\frac{1}{2n+3}\)
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4 tháng 10 2017 lúc 20:50
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}{nleft(n+1right)}right)frac{1}{3}2. cho M frac{1}{1.left(2n-1right)}+frac{1}{3.left(2n-3right)}+frac{1}{5.left(2n-5right)}+...+frac{1}{left(2n-3right).3}+frac{1}{left(2n-1right).1}N 1+frac{1}{3}+frac{1}{5}+...+frac{1}{2n-1}Rút gọn frac{M}{N}
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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 \(\frac{M}{N}\)
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 \(A=\frac{2n+1}{\left(n+1\right)^2}\)
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rút gọn
A= \(\frac{3}{\left(1.2^2\right)}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{n\left(n+1\right)^2}\)
Ta có:
\(2n+1=\left(n^2+2n+1\right)-n^2=\left(n+1\right)^2-n^2\Rightarrow\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
Thay vào ta rút gọn được các số hạng của A, cuối cùng được:
\(A=1-\frac{1}{\left(n+1\right)^2}\)
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Rút gọn biểu thức :
a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
Rút gọn: \(\frac{\left(\frac{-1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
Giúp mình nha các bạn :)
bai 1 :thuc hien phep tinh
a)\(\frac{-5}{2}:\left(\frac{3}{4}-\frac{1}{2}\right)\) b) \(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\) c)\(\left(\frac{-5}{24}+0,75+\frac{7}{12}\right):\left(-2\frac{1}{4}\right)^2\)
AI LAM NHANH VA DUNG THI MK SE TICK CHO NHA
MK DAG CAN GAP !!!!!!! CAC BAN LAM NHANH LEN NHA
Rút gọn biểu thúc
\(\frac{A}{B}=\frac{\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+\frac{1}{5\left(2n-5\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}}\)
\(\frac{1}{1.\left(2n-1\right)}+\frac{1}{3.\left(2n-3\right)}+...+\frac{1}{\left(2n-3\right).3}+\frac{1}{\left(2n-1\right).1}\)
\(=\frac{1}{2n}\left[\frac{2n-1+1}{1\left(2n-1\right)}+\frac{2n-3+3}{3\left(2n-3\right)}+...+\frac{3+2n-3}{\left(2n-3\right).3}+\frac{1+2n-1}{\left(2n-1\right).1}\right]\)
\(=\frac{1}{2n}\left(1+\frac{1}{2n-1}+\frac{1}{3}+\frac{1}{2n-3}+...+\frac{1}{2n-3}+\frac{1}{3}+\frac{1}{2n-1}+1\right)\)
\(=\frac{1}{n}\left(1+\frac{1}{3}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{1}{n}\).
Rút gọn biểu thức Sleft(xright)dfrac{1}{x^2}+dfrac{2}{x^3}+dfrac{3}{x^4}+...+dfrac{n}{x^{n+1}} bằng:A. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{n+1}left(x-1right)^2}B. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^{2n}left(x-1right)^2}C. Sdfrac{x^n-left(n+1right)x+n}{x^nleft(x-1right)^2}D. Sdfrac{x^{n+1}-left(n+1right)x+n}{x^nleft(x-1right)^2}
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Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:
A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)
C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)
