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Hướng dẫn giải:

Ta có phương trình quãng đường:  \( s=\frac{1}{2}a{{t}^{2}} \)

Do đó: \(\left\{ \begin{align}  & {{s}_{1}}=\frac{1}{2}a{{t}^{2}} \\  & {{s}_{2}}=\frac{1}{2}a.{{(2t)}^{2}}=\frac{4}{2}a{{t}^{2}} \\  & {{s}_{3}}=\frac{1}{2}a.{{(3t)}^{2}}=\frac{9}{2}a{{t}^{2}} \\  & …. \\  & {{s}_{n-1}}=\frac{1}{2}a.{{\left[ (n-1)t \right]}^{2}}=\frac{{{(n-1)}^{2}}}{2}a{{t}^{2}} \\  & {{s}_{n}}=\frac{1}{2}a.{{(nt)}^{2}}=\frac{{{n}^{2}}}{2}a{{t}^{2}} \\ \end{align} \right.\)

Suy ra: \(\left\{ \begin{align}  & \Delta {{s}_{1}}={{s}_{1}}=\frac{1}{2}a{{t}^{2}} \\  & \Delta {{s}_{2}}={{s}_{2}}-{{s}_{1}}=\frac{3}{2}a{{t}^{2}} \\  & \Delta {{s}_{3}}={{s}_{3}}-{{s}_{2}}=\frac{5}{2}a{{t}^{2}} \\  & … \\  & \Delta {{s}_{n}}={{s}_{n}}-{{s}_{n-1}}=\frac{1}{2}\left[ {{n}^{2}}-{{(n-1)}^{2}} \right]a{{t}^{2}}=\frac{2n-1}{2}a{{t}^{2}} \\ \end{align} \right.\)

Vậy:  \( \frac{\Delta {{s}_{2}}}{\Delta {{s}_{1}}}=3;\text{ }\frac{\Delta {{s}_{3}}}{\Delta {{s}_{1}}}=5;\text{ }…;\text{ }\frac{\Delta {{s}_{n}}}{\Delta {{s}_{1}}}=2n-1 \)