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AH
Akai Haruma
Giáo viên
23 tháng 5 2021
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.
NA
Ngoc Anh Thai
Giáo viên
28 tháng 3 2021
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
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
BT
5 tháng 12 2017
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)
9 tháng 5 2022
$\frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}(*)$
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
$\Rightarrow \frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}$
24 tháng 5 2018
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 :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
12 +22+32+...+n2
= 1.(2-1)+2.(3-1)+3.(4-1)+...+n.(n+1-1)
= (1.2+2.3+3.4+...+n.n(n+1)) - (1+2+3+...+n)
Dat A = 1.2+2.3+3.4+...+n.(n+1)
=> 3A = 1.2.3+2.3.3+3.4.3+...+n.(n+1).3
3A = 1.2.3+2.3(4-1)+3.4.(5-2)+...+n.(n+1).(n+2-n+1)
3A = (1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)) - (1.2.3+2.3.4+...+(n-1).n.(n+1))
3A = n.(n+1).(n+2)
\(\Rightarrow A=\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
ta co: 1+2+...+n = n.(n+1)/2
=> \(1^2+2^2+...+n^2=\frac{n.\left(n+1\right).\left(n+2\right)}{3}-\frac{n.\left(n+1\right)}{2}=\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
cop sai de hay sao z bn???
Sửa đề : 12 + 22 + 32 + ... + n2 = \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
VT <=> 1 ( 2 - 1 ) + 2 ( 3 - 1 ) + 3 ( 4 - 1 ) + ... + n [ ( n + 1 ) - 1 ]
= [ 1 . 2 + 2 . 3 + 3 . 4 + ... + n ( n + 1 ) ] - ( 1 + 2 + 3 + 4 + ... + n )
Đặt A = 1 . 2 + 2 . 3 + 3 . 4 + ... + n ( n + 1 ) . Ta có :
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 3n ( n + 1 )
=> 3A = 1.2.3 + 2.3 ( 4 - 1 ) + 3.4 ( 5 - 2 ) + ... + n ( n + 1 ) [ ( n + 2 ) - ( n - 1 ) ]
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n ( n + 1 ) ( n + 2 ) - ( n - 1 ) n ( n + 1 )
=> 3A = n ( n + 1 ) ( n + 2 )
=> A = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
=> VT = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)- ( 1 + 2 + 3 + 4 + ... + n )
= \(\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{\left(n+1\right)n}{2}\)
\(=\frac{2n\left(n+1\right)\left(n+2\right)-3n\left(n+1\right)}{6}\)
\(=\frac{n\left(n+1\right)\left(n+2\right)}{6}=VP\)( Đpcm )