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Tất cả các đẳng thức trên đều được chứng minh theo phương pháp quy nạp
Đặt n = k thì có đẳng thức
Chứng minh rằng n = k+1 cũng đúng ( vế trái (k+1) = vế phải (k+1) )
a) n2(n + 1) + 2n(n + 1)
= (n2 + 2n)(n + 1)
= n(n + 2)(n + 1) chia hết cho 6 vì là 3 số tự nhiên liên tiếp
b) (2n - 1)3 - (2n - 1)
= (2n - 1).[(2n - 1)2 - 1]
= (2n - 1).{ [ (2n - 1) + 1] . [ (2n - 1) -1 ] }
= *2n - 1) . 2n . (2n - 2) chia hết cho 8 vì là 3 số chẵn liên tiếp
c) (n + 2)2 - (n - 2)2
= n2 + 4n - 4 - (n2 - 4n + 4)
= n2 + 4n - 4 - n2 + 4n - 4
= 8n - 8 chia hết cho 8
Đặt \(T=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(2n-1\right)n}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{n}\)
\(=\frac{1}{2}-\frac{1}{n}< \frac{1}{2}^{\left(đpcm\right)}\) (không chắc nha)
Đặt \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{1}=\frac{1}{1},\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},....,\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=> \(A< \frac{1}{2^2}.\left[1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\right]\)
\(=\frac{1}{2^2}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2^2}.\left(2-\frac{1}{n+1}\right)=\frac{1}{2}-\frac{1}{4.\left(n+1\right)}\)
p/s: bài tớ ko bt đúng ko, nhưng tth bn làm vậy sẽ ko có quy luật, đoạn này
nếu cứ theo quy luật, tiếp tục sẽ ntn:\(\frac{1}{6^2}< \frac{1}{5.6};\frac{1}{8^2}< \frac{1}{6.7};\frac{1}{10^2}< \frac{1}{7.8}\)
Ta có:
\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)
\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)
\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)
\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)
...
\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)
\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)
=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
Xét trường hợp n chẵn:
\(1^2+2^2+3^2+...+n^2=\left(1^2+3^2+5^2+...+\left(n-1\right)^2\right)+\left(2^2+4^2+6^2+...+n^2\right)\)
\(=\frac{\left(n-1\right).n.\left(n+1\right)+n\left(n+1\right).\left(n+2\right)}{6}\)
\(=\frac{n\left(n+1\right).\left(n-1+n+2\right)}{6}\)
\(=\frac{n\left(n+1\right).\left(2n+1\right)}{6}\)
Tương tự với trường hợp n lẻ . ta có \(\text{ĐPCM}\)
\(A=1^2+2^2+3^2+....+n^2\)
\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+n\left[\left(n+1\right)-1\right]\)
\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n\)
\(=\left[1.2+2.3+3.4+....+n\left(n+1\right)\right]-\left(1+2+3+....+n\right)\)
Ta có :
\(1.2+2.3+3.4+....+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)(cái này tự CM nha)
\(1+2+3+....+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(đpcm)