Question

10. CMR:

\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\) = n

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Answer 1

\(A=\sqrt[]{1+2+3+...+\left(n-1\right)+n+...+3+2+1}\)

Ta có :

\(1+2+3+...+\left(n-1\right)=\left(n-1\right)+...+3+2+1=\left[\left(n-1\right)-1\right]+1\left(n-1+1\right):2\)

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

\(\Rightarrow A=\sqrt[]{\dfrac{\left(n-1\right)n}{2}.2+n}\)

\(\Rightarrow A=\sqrt[]{\left(n-1\right)n+n}\)

\(\Rightarrow A=\sqrt[]{n^2-n+n}\)

\(\Rightarrow A=\sqrt[]{n^2}\)

\(\Rightarrow A=n\left(n>0\right)\)

\(\Rightarrow dpcm\)

Answer 2

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