Question

Cho : \(a_n=1+2+3+....+n\)

a) Tính \(a_n+1\)

b) CM: \(a_n+a_{n+1}\) là số chính phương

Answer 1

a) \(a_n+1=\left(1+2+3+...+n\right)+1=\dfrac{n\left(n+1\right)}{2}+1\)

b) Ta có:

\(a_n+a_{n+1}=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n+1\right)\left(n+2\right)}{2}=\dfrac{n\left(n+1\right)+\left(n+1\right)\left(n+2\right)}{2}=\dfrac{\left(n+1\right)\left(2n+2\right)}{2}=\left(n+1\right)^2\)

Vậy a_n + a_{n + 1} là số chính phương

Answer 2

a,

\(a_n+1=1+2+...+n+1=\dfrac{n\left(n+1\right)}{2}+1=\dfrac{n\left(n+1\right)+2}{2}\)

b,

\(a_n+a_{n+1}=2a_n+n+1\)

\(=\dfrac{n\left(n+1\right)}{2}\cdot2+n+1=n\left(n+1\right)+\left(n+1\right)\)

\(=\left(n+1\right)^2\) là số chính phương