问题
解答题
将边长分别为
|
答案
S2-S1=(2
)2-(2
)2=2×(22-12)=6;2
S3-S2=(3
)2-(22
)2=2×(32-22)=10;2
S4-S3=(4
)2-(32
)2=2×(42-32)=14;2
…
Sn-Sn-1=(n
)2-[(n-1)2
]2=2×[n2-(n-1)2]=4n-2.2
故答案为:4n-2.
将边长分别为
|
S2-S1=(2
)2-(2
)2=2×(22-12)=6;2
S3-S2=(3
)2-(22
)2=2×(32-22)=10;2
S4-S3=(4
)2-(32
)2=2×(42-32)=14;2
…
Sn-Sn-1=(n
)2-[(n-1)2
]2=2×[n2-(n-1)2]=4n-2.2
故答案为:4n-2.