Question

Cho \(a_n=1+2+...+n\)

a) Tính \(a_{n+1}\)

b) CMR \(a_n+a_{n+1}\)là một số chính phương.

Answer 1

a,Ta có : a_n+1=1+2+....+n+(n+1)

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

b,Ta lại có :\(\Rightarrow a=\frac{\left(n+1\right)\left[\left(n-1\right):1+1\right]}{2}=\frac{\left(n+1\right)\left(n\right)}{2}\)

\(\Rightarrow a_n+a_{n+1}=\frac{\left(n+2\right)\left(n+1\right)}{2}+\frac{\left(n+1\right)n}{2}\)

\(\Rightarrow a_n+a_{n+1}=\frac{\left(n+1\right)\left[\left(n+2\right)+n\right]}{2}=\frac{\left(n+1\right)\left(2n+2\right)}{2}\)

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

=>ĐPCM