bạn làm xong bài này chưa dạy mình với

giup giai cau nay voi

:$\frac{n(n+2)}{(n+1)^2}

                                          =1-\frac{1}{(x+1)^2}

                                          > 1-\frac{1}{x(x+2)}

                                          = 1-\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})$

Thay lần lượt vô

\(\dfrac{n^2-1}{n^2}=1-\dfrac{1}{n^2}>1-\dfrac{1}{\left(n-1\right)n}\)

Từ đó ta có:

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{50^2-1}{50^2}>1-\dfrac{1}{1.2}+1-\dfrac{1}{2.3}+...+1-\dfrac{1}{49.50}\)

\(\Rightarrow A>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A>49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\\ A=\left(1+1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\\ A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)\)

Có \(\dfrac{1}{4}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\\ \dfrac{1}{9}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\\ \dfrac{1}{16}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\\ ...\\ \dfrac{1}{2500}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1-\dfrac{1}{50}< 1\\ \Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}< 1\)

\(\Rightarrow A=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{2500}\right)>49-1\\ \Rightarrow A>48\)

a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)

= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)

= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)

b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)

= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)

= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)

a) S = 1.2 + 2.3 + 3.4 + ... + 99.100

S có thể được viết lại thành:

S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)

= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98

= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)

Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:

S = n(n+1)(2n+1)/6

Với n = 99, ta có:

S = 99.100.199/6 = 331650

Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:

S = n(n+1)/2

Với n = 98, ta có:

S = 98.99/2 = 4851

Do đó, S = 331650 - 4851 = 326799

b) B = 4924.12517.28−530.749.45529.162.748

B có thể được viết lại thành:

B = (4924.12517.28) / (530.749.45529.162.748)

B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)

B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529

B = 108 / 45529

c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101

C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101

C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)

C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)

C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)

d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018

D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^