Chứng minh các mệnh đề sau
\(a,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{n}{n+1}\) \(\forall n\in N\) *
\(b,1+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\forall n\ge2\)
a: \(VT=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)
chứng tỏ rằng với mọi n thuộc N ta luôn có
\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
Chứng minh các mệnh đề sau theo phương pháp qui nạp dãy số:
\(a,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{n}{n+1}\) \(\forall n\in N\) *
\(b,1+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\forall n\ge2\)
Chứng minh các mệnh đề sau theo phương pháp qui nạp dãy số:
\(a,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{n}{n+1}\forall n\in N\)*
\(b,1+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\forall n\ge2\)
\(a,n=1\Leftrightarrow\dfrac{1}{1.2}=\dfrac{1}{2}\left(đúng\right)\\ G\text{/}s:n=k\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{k\left(k+1\right)}=\dfrac{k}{k+1}\\ \text{Với }n=k+1\\ \text{Cần cm: }\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{k\left(k+1\right)}+\dfrac{1}{\left(k+1\right)\left(k+2\right)}=\dfrac{k+1}{k+2}\\ \text{Ta có }VT=\dfrac{k}{k+1}+\dfrac{1}{\left(k+1\right)\left(k+2\right)}=\dfrac{k^2+2k+1}{\left(k+1\right)\left(k+2\right)}\\ =\dfrac{\left(k+1\right)^2}{\left(k+1\right)\left(k+2\right)}=\dfrac{k+1}{k+2}=VP\)
Vậy với \(n=k+1\) thì mệnh đề cũng đúng
Vậy theo pp quy nạp ta đc đpcm
chứng minh rằng \(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)
Chứng minh \(\forall\) n \(\in\) N, n > 1 ta có \(\dfrac{1}{n-1}-\dfrac{1}{n}>\dfrac{1}{n^2}>\dfrac{1}{n}-\dfrac{1}{n+1}\)
Chứng minh các mệnh đề sau:
\(a,1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\) \(\forall n\in N\) *
\(b,1.2+2.3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\) \(\forall n\in N\) *
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
Chứng minh rằng :
a) \(\dfrac{1.3.5.....39}{21.22.23.....40}=\dfrac{1}{2^{20}}\)
b) \(\dfrac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\dfrac{1}{2^n}\) với \(n\in\) N*
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
