Xét trường hợp n chẵn:

\(1^2+2^2+3^2+...+n^2=\left(1^2+3^2+5^2+...+\left(n-1\right)^2\right)+\left(2^2+4^2+6^2+...+n^2\right)\)

\(=\frac{\left(n-1\right).n.\left(n+1\right)+n\left(n+1\right).\left(n+2\right)}{6}\)

\(=\frac{n\left(n+1\right).\left(n-1+n+2\right)}{6}\)

\(=\frac{n\left(n+1\right).\left(2n+1\right)}{6}\)

Tương tự với trường hợp n lẻ . ta có \(\text{ĐPCM}\)

\(A=1^2+2^2+3^2+....+n^2\)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+n\left[\left(n+1\right)-1\right]\)

\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n\)

\(=\left[1.2+2.3+3.4+....+n\left(n+1\right)\right]-\left(1+2+3+....+n\right)\)

Ta có :

\(1.2+2.3+3.4+....+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)(cái này tự CM nha)

\(1+2+3+....+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(đpcm)

Cảm ơn 2 bạn nha

Tất cả các đẳng thức trên đều được chứng minh theo phương pháp quy nạp

Đặt n = k thì có đẳng thức

Chứng minh rằng n = k+1 cũng đúng ( vế trái (k+1) = vế phải (k+1) )

thi giai ra luon dj

Tất cả các đẳng thức trên đều được chứng minh theo phương pháp quy nạp

Đặt n = k thì có đẳng thức

Chứng minh rằng n = k+1 cũng đúng ( vế trái (k+1) = vế phải (k+1) )

bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)

Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)

\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)

\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)

\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)

\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)

\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)

Vì m+n+p=0=>m+n=-p

\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)

\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)

\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)

\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)

\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)

\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)

\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)

\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)

\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)

\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)

\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)

Từ (1),(2),(3) suy ra :

\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)

\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)

*Tới đây để tính được m³+n³+p³,ta cần CM được bài toán phụ sau:

Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)

Từ m+n+p=0=>m+n=-p

Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)

\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)

Vậy ta đã CM được bài toán phụ

*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)

Vậy A=9

bài 2)

a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:

\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)

\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)

suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)

Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)

\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)

...........................

\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)

\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)

\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)

Vậy A=2036/37

b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 1⁴=1 mà

Nhận thấy các thừa số của B có dạng tổng quát:

\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)

\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)

\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)

Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)

Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)

Vậy B=1/221

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 :

\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)

\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)

a) A = 1² + 2² + ...+ n² = 1.(2 - 1) + 2.(3 - 1) + ...+ n.(n+ 1 - 1) = [1.2 + 2.3 + ...+ n.(n+1)] - (1 + 2 + ... + n)

Tính B = 1.2 + 2.3 + ...+ n.(n+1)

=> 3.B = 1.2.3 + 2.3.3 +3.4.3 + ...+ n.(n+1).3

= 1.2.3 + 2.3.(4 -1) + 3.4 .(5 - 2) + ...+ n.(n+1).((n+2) - (n-1) )

= [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)] = n(n+1)(n+2)

=> B = n(n+1).(n+2)/3

Tính 1 + 2 + 3 + ..+ n =(n+1).n / 2

Vậy A =  n(n+1).(n+2)/3 - (n+1).n / 2 = n(n+1).(2n+1) / 6

Ta có: \(n^3=n.n.n=n.\left(\frac{n+1+n-1}{2}\right).n\left(\frac{\left(n+1\right)-\left(n-1\right)}{2}\right)\)

\(=\left(\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)}{2}\right).\left(\frac{n\left(n+1\right)}{2}-\frac{n\left(n-1\right)}{2}\right)=\left(\frac{n\left(n+1\right)}{2}\right)^2-\left(\frac{n\left(n-1\right)}{2}\right)^2\)

(Áp dụng công thức a² - b² = (a-b).(a+b))

Áp dụng vào ta có: \(1^3=\left(\frac{1.2}{2}\right)^2-\left(\frac{1.0}{2}\right)^2\)

                             \(2^3=\left(\frac{2.3}{2}\right)^2-\left(\frac{2.1}{2}\right)^2\)

                             \(3^3=\left(\frac{3.4}{2}\right)^2-\left(\frac{3.2}{2}\right)^2\)

                            ......................

                            \(n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2-\left(\frac{n\left(n-1\right)}{2}\right)^2\)

Cộng từng vế ta được:

\(1^3+2^3+....+n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2\)

tính hộ cái nha

S=5/1*6+5/6*11+5/11*16+5/16*21+...+5/96*101+5/101*106

a)A=n/n+1=n/n+0/1

   B=n+2/n+3=n/n  +  2/3

ta có:0<2/3

=>A<B