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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.
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
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
$\frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}(*)$
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
$\Rightarrow \frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}$
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)
Áp dụng phương pháp quy nạp:
- Với n=1; có (1+1)=2 chia hết cho 21
- Giả sử với n=k thì (k+1)(k+2)...2k chia hết cho 2k
CM: (k+1+1)(k+1+2....(2k+1) chia hết cho 2k+1
Ta có: (k+1+1(k+1+2)...(2k+1)=(k+2)(k+3)...2k.2(k+1)=2(k+1)(k+2)...2k chia hết cho 2.2k =2k+1
Vậy (n+1)(n+2)(n+3)...2n chia hết cho 2n, thương là q
\(\Rightarrow p=\frac{\left(n+1\right)\left(n+2\right)....2n}{2^n}=\frac{1.2...n\left(n+1\right).\left(n+2\right)...2n}{1.2...n.2^n}=\frac{\left(2n\right)!}{n!.2n}\)
Chứng minh bằng phương pháp quy nạp
với n=1=> ( 1+1) = 2 chia hết cho 21
giả sử với n=k thì (k+1)(k+2). ... . 2k chia hết cho 2k
cần chứng minh: (k+1+1)(k+1+2). ... . 2(k+1) chia hết cho 2k+1
Ta có:(k+1+1)(k+1+2). ... . 2(k+1)=(k+2)(k+3). ... . 2k.2(k+1)=2(k+1).(k+2). ... .2k chia hết cho 2 . 2k = 2k+1
Vậy (n+1)(n+2). ... .2n chia hết cho 2n
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
Nếu n= 2k+ 1 \((k\in N)\)
Thì n+1= (2k+1)+1=\((2k+2)⋮2^n\)
Nếu n=2k\(\left(k\in N\right)\)
Thì n+ 2=( 2k+2)\(⋮\)\(2^n\)...
vậy \(\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n⋮2^n\)