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1.A = 21 + 22 + 23 + 24 + ... + 259 + 260
Xét .dãy số: 1; 2; 3; 4; .... 59; 60 Dãy số này có 60 số hạng vậy A có 60 hạng tử.
vì 60 : 2 = 30 nên nhóm hai số hạng liên tiếp của A vào một nhóm thì ta được:
A = (21 + 22) + (23 + 24) +...+ (259 + 260)
A = 2.(1 + 2) + 23.(1 +2) +...+ 259.(1 +2)
A =2.3 + 23.3 + ... + 259.3
A =3.( 2 + 23+...+ 259)
Vì 3 ⋮ 3 nên A = 3.(2 + 23 + ... + 259)⋮3 (đpcm)
# Mik làm ý A trước nhé, mik sợ dài :
- Với n = 1 \(\Rightarrow1=\frac{1.2.3}{6}\)( đúng )
- Giả sử đẳng thức cũng đúng với\(n=k\)hay :
\(1^2+2^2+3^2+...+k^2=\)\(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Ta cần chứng minh nó cũng đúng với\(n=k+1\)hay :
\(1^2+2^2+3^2+...+k^2+\left(k+1\right)^2=\)\(\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\)
Thật vậy, ta có:
\(1^2+2^2+3^2+...+k^2+\left(k+1\right)^2=\)\(\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(\Rightarrow\left(k+1\right)\left(\frac{k\left(2k+1\right)}{6}+k+1\right)=\)\(\left(k+1\right)\left(\frac{2k^2+k+6k+6}{6}\right)\)
\(\Rightarrow\)\(\left(k+1\right)\left(\frac{2k^2+7k+6}{6}\right)=\)\(\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)( đpcm )
# giờ mik làm ý B nha !
- Với n = 1 \(\Rightarrow\)1 = 1 ( đúng )
Giả sử bài toán đúng với\(n=k\left(n\inℕ^∗\right)\)thì ta có :
1 + 23 + 33 + .... + k3 = \(\left[\frac{n\left(n+1\right)}{2}\right]^2\left(1\right)\)
Ta cần chứng minh đề bài đúng với\(n=k+1\)tức là :
13 + 23 + 33 + ...... + n3 = \(\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\left(2\right)\)
Đặt \(B=1^3+2^3+...+\left(k+1\right)^3\)
\(=\left(\frac{k\left(k+1\right)}{2}\right)^2+\left(k+1\right)^3\)theo ( 1 )
\(=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)theo ( 2 )
\(\Rightarrow\left(1\right),\left(2\right)\)đều đúng
Mà \(\left[\frac{n\left(n+1\right)}{2}\right]^2=\)\(\frac{n^2\left(n+1\right)^2}{4}\)
\(\Rightarrow\)\(1^3+2^3+...+n^3=\)\(\frac{n^2\left(n+1\right)^2}{4}\)( đpcm )
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.
\(1^2+2^2+...+n^2=1+2\left(1+1\right)+...+n\left(n-1+1\right)=1+2+1.2+3+2.3+...+n+\left(n-1\right)n\)
\(=\left(1+2+3+...+n\right)+\left[1.2+2.3+...+\left(n-1\right)n\right]=\dfrac{\left(n+1\right)\left(\dfrac{n-1}{1}+1\right)}{2}+\dfrac{1.2.3+2.3.3+...+\left(n-1\right)n.3}{3}=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3+2.3.\left(4-1\right)+...+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{1.2.3-1.2.3+2.3.4-...-\left(n-2\right)\left(n-1\right)n+\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{3n\left(n+1\right)+2\left(n-1\right)n\left(n+1\right)}{6}=\dfrac{2n^3+3n^2+n}{6}=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n=\dfrac{1}{3}n\left(n^2+\dfrac{3}{2}n+\dfrac{1}{2}\right)=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)\)
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
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
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)
Ta thấy:
\(\sqrt{1^3}=1\)
\(\sqrt{1^3+2^3}=\sqrt{9}=3=1+2\)
\(\sqrt{1^3+2^3+3^3}=\sqrt{36}=6=1+2+3\)
..........................................
=>\(\sqrt{1^3+2^3+...+n^3}=1+2+...+n\Rightarrow1^3+2^3+...+n^3=\left(1+2+...+n\right)^2\)