Question

[pham nhu nguyen]

a) cho A = 1+3+5+7+...+(2n+1) Voi n thuoc N

chung to rang A la so chinh phuong

b)B=2+4+6+8+...+2n voi n thuocN

so B co phai la so chinh phuong ko

Answer 1

[shitbo]

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

Answer 2

[Xyz OLM]

A = 1 + 3 + 5 + 7 + ... + 2n + 1

   = \(\left[\left(2n+1-1\right):2+1\right].\left(\frac{2n+1+1}{2}\right)\)

   = \(\left(n+1\right).\left(n+1\right)\)

   = \(\left(n+1\right)^2\)

=> A là số chính phương (đpcm)

b) \(2+4+6+...+2n\)

= \(\left[\left(2n-2\right):2+1\right].\frac{2n+2}{2}\)

= \(n.\left(n+1\right)\)

= \(n^2+n\)

\(\Rightarrow\)B không là số chính phương