Question

Cho \(a_n=1+2+3+...+n\). Chứng minh rằng \(a_n+a_{n+1}\) là một số chính phương.

Answer 1

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

\(\Rightarrow a_{n+1}=1+2+3+...+n+\left(n+1\right)=\dfrac{\left(n+1\right)\left(n+2\right)}{2}\)

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

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

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

\(\Rightarrow dpcm\)

Answer 2

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