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Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$\frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}(*)$
Với $n=1$ thì $(*)\Leftrightarrow \frac{1}{2}=\frac{1}{2}$
Vậy $(*)$ đúng với $n=1$
Giả sử với $n=k$,$ k\in \mathbb{N^*}$ thì $(*)$ đúng, tức là:
$\frac{1.3.5...(2k-1)}{(k+1)(k+2)...(k+k)}=\frac{1}{2^k}$
Ta cần chứng minh với $n=k+1$ thì $(*)$ đúng, tức là:
$\frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1}{2^{k+1}}=\frac{1}{2^k}.\frac{1}{2}$
$\Leftrightarrow \frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...(k+k)}$
$\Leftrightarrow \frac{1.3.5...(2k-1)2k(2k+1)}{(k+2)(k+3)...2k(2k+1)(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...2k}$
$\Leftrightarrow \frac{2k(2k+1)}{2k(2k+1)(2k+2)}=\frac{1}{2(k+1)}$
$\Leftrightarrow \frac{1}{(2k+2)}=\frac{1}{2(k+1)}$
Do đó với $n=k+1$ thì $(*)$ đúng
$\Rightarrow \frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1^2+2^2+3^2...+n^2=1+2\left(1+1\right)+3\left(2+1\right)+...+n\left(n-1+1\right)\\ =1+1\cdot2+2+3\cdot2+3+...+n\left(n-1\right)+n\\ =\left(1+2+3+...+n\right)+\left[1\cdot2+2\cdot3+...+n\left(n-1\right)\right]\)
Ta có \(1\cdot2+2\cdot3+...+n\left(n-1\right)\)
\(=\dfrac{1}{3}\left[1\cdot2\cdot3+2\cdot3\cdot3+...+3n\left(n-1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+n\left(n-1\right)\left(n+2+n+1\right)\right]\\ =\dfrac{1}{3}\left(1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)\right)\\ =\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\)
\(\Rightarrow1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\\ =\dfrac{3n\left(n+1\right)+2n\left(n-1\right)\left(n+1\right)}{6}=\dfrac{n\left(n+1\right)\left(3+2n-2\right)}{6}\\ =\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(M=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2).(n+1)(n+2)....2n}=\frac{(2n-1)!}{2.1.2.2.2.3...2(n-1).(n+1).(n+2)...2n}\)
\(=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).(n+1).(n+2)....2n}=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).n(n+1)..(2n-1).2}\)
\(=\frac{(2n-1)!}{2^{n-1}.(2n-1)!.2}=\frac{1}{2^{n-1}.2}<\frac{1}{2^{n-1}}\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1.2.3 + 2.3.4 + 3.4.5 ... + n(n + 1)(n + 2)
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + n(n + 1)(n + 2).4
4A = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2)+ ... + n(n + 1)(n + 2)[(n + 3) - (n - 1)]
4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + n(n + 1)(n + 2)(n + 3) - (n-1)n(n+1)(n+2)
4A = n(n+1)(n+2)(n+3)
A = n(n + 1)(n+2)(n + 3) : 4
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(1.3.5.7.9...\left(2n-1\right)=\frac{\left[1.3.5.7.9....\left(2n-1\right)\right].\left[2.4.6.8...2n\right]}{2.4.6.8....2n}=\frac{1.2.3.4.5.6....2n}{\left(2.1\right).\left(2.2\right).\left(2.3\right)\left(2.4\right)....\left(2.n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{1.2.3.4.5.6....2n}{\left(2.2.2.....2\right).\left(1.2.3.4.....n\right)}=\frac{\left(1.2.3.4.....n\right)\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}{2^n.\left(1.2.3.4....n\right)}\)
=> \(1.3.5.7.9...\left(2n-1\right)=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n}\)
=> \(\frac{1.3.5.7.9...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}=\frac{1}{2^n}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(n-1\right)\left(n+1\right)\left(n^2\right)\left(n^2+1\right)\)
\(A=\left(n-1\right)n\left(n+1\right).n\left(n^2+1\right)\left(I\right)\)
\(A=\left[\left(n-1\right)\left(n+1\right).n^2\right]\left(n^2-4+5\right)\)
\(=\left(n-1\right)\left(n+1\right).n^2\left(n^2-2^2\right)+5\left(n-1\right)\left(n+1\right).n^2\)
\(=\left(n-1\right)\left(n+1\right).n^2\left(n-2\right)\left(n+2\right)+5\left(n-1\right)\left(n+1\right).n^2\)
\(=\left(n-2\right)\left(n-1\right)\left(n+1\right)\left(n+2\right).n^2+5\left(n-1\right)\left(n+1\right).n^2\left(II\right)\)
1)với (I) A là tích của 3 số tự nhiên liên tiếp => chia hết cho 2 &3
2) với bửu thức (II) A là tổng hai số hạng
số hạng đầu là tích của 5 số tự nhiên liên tiếp=> chia hết cho 5
số hạng sau hiển nhiên chia hết cho 5 do có thừa số 5
KL
Với (I) A chia hết cho 2&3
Với (II) A chia hết cho 5
(I)&(II)=> điều bạn muốn tìm
n2(n+3)-(n+3)=n2(n+3)-1(n+3)=(n2-1)(n+3)
=>đpcm
đpcm có phải là điều phải chứng minh ko