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\(1^2+2^2+3^2+.......+n^2=1\times\left(2-1\right)+2\times\left(3-1\right)+.......+n\left(\left(n+1\right)-1\right)\)=\(\left(1.2+2.3+3.4+......+n\left(n+1\right)\right)-\left(1+2+3+.....+n\right)\)=\(\frac{n\left(n+1\right)\left(n+2\right)-0.1.2}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
sử dụng qui nạp:
1² + 2² + 3² + 4² + ...+ n² = \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) (*)
(*) đúng khi n= 1
giả sử (*) đúng với n= k, ta có:
1² + 2² + 3² + 4² + ...+ k² = \(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)
ta cm (*) đúng với n = k +1, thật vậy từ (1) cho ta:
1² + 2² + 3² + 4² + ...+ k² + (k + 1)² = \(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) + (k + 1)²
= (k+1)\(\left(\frac{k\left(2k+1\right)}{6}+\left(k+1\right)\right)\)= (k + 1)\(\frac{2k^2+k+6k+6}{6}\)
= (k + 1)\(\frac{2k^2+7k+6}{6}\) = (k + 1)\(\frac{2k^2+4k+3k+6}{6}\)
= (k + 1)\(\frac{2k\left(k+2\right)+3\left(k+2\right)}{6}\) = (k + 1)\(\frac{\left(k+2\right)\left(2k+3\right)}{6}\)
vậy (*) đúng với n = k + 1, theo nguyên lý qui nạp (*) đúng với mọi n thuộc N*
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
Đặt \(A_k=1+2+3+4+.....+k=\frac{k\left(k+1\right)}{2}\Rightarrow A_k^2=\frac{k^2\left(k+1\right)^2}{4}\)
\(A_{k-1}=1+2+3+4+.....+\left(k-1\right)=\frac{k\left(k-1\right)}{2}\Rightarrow A_{k-1}^2=\frac{k^2\left(k-1\right)^2}{4}\)
\(\Rightarrow A_k^2-A_{k-1}^2=\frac{k^2\left(k+1\right)^2-k^2\left(k-1\right)^2}{4}=\frac{k^2\left(k^2+2k+1-k^2+2k-1\right)}{4}=\frac{4k^3}{4}=k^3\)
Khi đó:
\(1^3=A_1^2\)
\(2^3=A_2^2-A_1^2\)
\(3^3=A_3^2-A_2^2\)
\(.........................................................................................\)
\(n^3=A_n^2-A_{n-1}^2\)
\(\Rightarrow1^3+2^3+3^3+.....+n^3=A_n^2=\left(1+2+3+......+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\)
Đề ghi sót . Vế cuối là móc vuông đó bình phương chư
Ta có :
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
\(A< \frac{1}{4}-\frac{1}{4n}\)
Lại có \(n>0\) nên \(\frac{1}{4n}>0\)
\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)
Đặt A = 12 + 22 + 32 + 42 + ... + n2
A = 1 + (1 + 1).2 + (1 + 2).3 + (1 + 3).4 + ... + (1 + n - 1).n
A = 1 + 1.2 + 2 + 3 + 2.3 + 4 + 3.4 + ... + n + (n - 1).n
A = [1.2 + 2.3 + 3.4 + ... + (n - 1).n] + (1 + 2 + 3 + 4 + ... + n)
A = [1.2 + 2.3 + 3.4 + ... + (n - 1).n] + \(\frac{n.\left(n+1\right)}{2}\)
Đặt B = 1.2 + 2.3 + 3.4 + ... + (n - 1).n
3B = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + (n - 1).n.[(n + 1) - (n - 2)]
3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + (n - 1).n.(n + 1) - (n - 2).(n - 1).n
3B = (n - 1).n.(n + 1)
\(B=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)
\(A=\frac{n.\left(n+1\right)}{2}+\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)
\(A=\frac{3n.\left(n+1\right)+2.\left(n-1\right).n.\left(n+1\right)}{6}\)
\(A=\frac{n.\left(n+1\right).\left(3+2n-2\right)}{6}=\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\left(đpcm\right)\)
Đề là chứng minh N < 1/4 sẽ đúng hơn
Ta có :
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(\Rightarrow2^2.N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
Ta lại có :
\(4N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}\)
\(\Rightarrow N< \left(1-\frac{1}{n}\right):4=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
Mà \(n\in N;n\ge2\)=> 1 -\(\frac{1}{n}\)< 1
=> \(N< \frac{1}{4}\left(1-\frac{1}{n}\right)< \frac{1}{4}\)
=> \(N< \frac{1}{4}\)( đpcm )
1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)
=> còn lại thì bạn có thể tự chứng minh
ĐẶT: \(A=1^2+2^2+3^2+....+n^2\)
\(=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+.....+n.\left(n+1-1\right)\)
\(=1.2-1+2.3-2+3.4-3+.....+n.\left(n-1\right)-n\)
\(=\left[1.2+2.3+3.4+....+n.\left(n+1\right)\right]-\left(1+2+3+...+n\right)\)
\(=\frac{n.\left(n+1\right).\left(n+2\right)}{3}-\frac{n.\left(n+1\right)}{2}\)
\(=n.\left(n+1\right).\left(n+\frac{2}{3}-\frac{1}{2}\right)\)
= \(n.\left(n+1\right).\left(\frac{2n+4}{3}-\frac{1}{2}\right)\)
\(=n.\left(n+1\right).\frac{2n+4-3}{6}\)
\(=\frac{n.\left(n+1\right).\left(2n+1\right)}{6}\)
Đặt \(M=1^2+2^2+3^2+...+n^2\)
\(M=1.1+2.2+3.3+...+n.n\)
\(M=\left(0+1\right)1+\left(1+1\right)2+\left(2+1\right)3+...+\left(n-1+1\right)n\)
\(M=0.1+1.1+1.2+1.2+2.3+1.3+...+\left(n-1\right)n+1.n\)
\(M=\left(0.1+1.2+2.3+...+\left(n-1\right)n\right)+\left(1.1+1.2+1.3+...+1.n\right)\)
\(M=\left(1.2+2.3+...+\left(n-1\right)n\right)+\left(1+2+3+...+n\right)\)
Đặt A=(2.3+3.4+...+(n-1)n và B=1+2+3+...+n rồi tự chứng minh được