A =2^2-1^2/1^2.2^2 + 3^2-2^2/2^2.3^2 + ..... + 2016^2-2015^2/2015^2.2016^2

= 1/1^2-1/2^2+1/2^2-1/3^2+.....+1/2015^2-1/2016^2
= 1-1/2016^2 < 1

=> ĐPCM

k mk nha

Mk hơi bối rối,bn dùng cái gõ phương trình trên thanh công cụ được ko.

Ta có: \(\frac{3}{1^2.2^2}=\frac{1}{1^2}-\frac{1}{2^2}\); \(\frac{5}{2^2.3^2}=\frac{1}{2^2}-\frac{1}{3^2}\); \(\frac{7}{3^2.4^2}=\frac{1}{3^2}-\frac{1}{4^2}\);....; \(\frac{4031}{2015^2.2016^2}=\frac{1}{2015^2}-\frac{1}{2016^2}\)

=> \(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2015^2}-\frac{1}{2016^2}\)

=> \(A=1-\frac{1}{2016^2}< 1\)

=> A < 1

Câu f:

F = 1.2 + 2.3 + 3.4 + ... + n(n + 1)

F = 1[1.2.3 + 2.3.3 + 3.4.3 + ... + n.(n+1).3].1/3

F = [1.2.3 +2.3.(4-1) + 3.4.(5-2) +..+n(n+1).(n+2-n-1)].1/3

F = [1.2.3+2.3.4-1.2.3+...+n(n+1)(n+2)-(n-1).n.(n+1)].1/3

F= n.(n+1).(n+2)/3

g) G= 1.2.3+2.3.4+3.4.5+...+99.100.101

4G =1.2.3.4 + 2.3.4.4 + ...+99.100.101.4

4G =1.2.3.4 + 2.3.4.(5-1) + ...+99.100.101.(102-98)

4G = 1.2.3.4 +2.3.4.5- 1.2.3.4+...+99.100.101.102-98.99.100.101

4G = 99.100.101.102

G = 99.100.101.102/4

bó tay kkk

B = 6 + 6^3 + 6^5 + ... + 6^2015

=> 6^2.B = 6^2(6 + 6^3 + 6^5 + ... + 6^2015

=> 36B = 6^2.6 + 6^3.6 + 6^5.6 + ... + 6^2015 .6 

=> 36B = 6^3 + 6^4 + 6^6 + ... + 6^2016

Lấy 36B trừ đi B, ta có:

     35B = 6^2016 - 6 

=> B = (6^2016 - 6)/35