Question

[Trần Nhật Quang]

cm với mọi số nguyên dương,ta có

2^2+5^2+8^2+...+(3n-1)^2=n(6n^2+3n-1)/2

Answer 1

[Rimuru tempest]

\(2^2+5^2+8^2+...+\left(3n-1\right)^2=\dfrac{n\left(6n^2+3n-1\right)}{2}\left(1\right)\)

Với n=1

\(VT=4;VP=4\)

(1) đúng với n=1

Giả sử (1) đúng với n=\(k\ge1\)

\(2^2+5^2+8^2+...+\left(3k-1\right)^2=\dfrac{k\left(6k^2+3k-1\right)}{2}\)

Ta cần phải chứng minh (1) đúng với n=k+1

\(\Leftrightarrow2^2+5^2+8^2+...+\left(3k-1\right)^2+\left[3\left(k+1\right)-1\right]^2=\dfrac{\left(k+1\right)\left[6\left(k+1\right)^2+3\left(k+1\right)-1\right]}{2}\)

\(\Leftrightarrow2^2+5^2+8^2+...+\left(3k-1\right)^2+\left(3k+2\right)^2=\dfrac{\left(k+1\right)\left(6k^2+15k+8\right)}{2}\)

\(VT=\dfrac{k\left(6k^2+3k-1\right)}{2}+\left(3k+2\right)^2=\dfrac{6k^3+3k^2-k+18k^2+24k+8}{2}\)

\(=\dfrac{6k^3+21k^2+23k+8}{2}=\dfrac{6k^3+15k^2+8k+6k^2+15k+8}{2}\)

\(=\dfrac{k\left(6k^2+15k+8\right)+\left(6k^2+15k+8\right)}{2}=\dfrac{\left(6k^2+15k+8\right)\left(k+1\right)}{2}\)

\(\Leftrightarrow VT=VP\)

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