题目内容
【题目】观察下列等式:
(sin )﹣2+(sin
)﹣2=
×1×2;
(sin )﹣2+(sin
)﹣2+(sin
)﹣2+sin(
)﹣2=
×2×3;
(sin )﹣2+(sin
)﹣2+(sin
)﹣2+…+sin(
)﹣2=
×3×4;
(sin )﹣2+(sin
)﹣2+(sin
)﹣2+…+sin(
)﹣2=
×4×5;
…
照此规律,
(sin )﹣2+(sin
)﹣2+(sin
)﹣2+…+(sin
)﹣2= .
【答案】n(n+1)
【解析】解:观察下列等式:(sin )﹣2+(sin
)﹣2=
×1×2;(sin
)﹣2+(sin
)﹣2+(sin
)﹣2+sin(
)﹣2=
×2×3;(sin
)﹣2+(sin
)﹣2+(sin
)﹣2+…+sin(
)﹣2=
×3×4;(sin
)﹣2+(sin
)﹣2+(sin
)﹣2+…+sin(
)﹣2=
×4×5;
…
照此规律(sin )﹣2+(sin
)﹣2+(sin
)﹣2+…+(sin
)﹣2=
×n(n+1),
故答案为: n(n+1)
由题意可以直接得到答案.;本题考查了归纳推理的问题,关键是找到相对应的规律,属于基础题.
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