1. 看图找规律,再填空。
① $ 2 = 1 × 2 $
② $ 2 + 4 = 2 × 3 $
③ $ 2 + 4 + 6 = $
④ $ 2 + 4 + 6 + 8 = $
根据上面的规律写一写。
$ 2 + 4 + 6 + 8 + 10 = $
$ 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = $
① $ 2 = 1 × 2 $
② $ 2 + 4 = 2 × 3 $
③ $ 2 + 4 + 6 = $
3
$ × $4
④ $ 2 + 4 + 6 + 8 = $
4
$ × $5
根据上面的规律写一写。
$ 2 + 4 + 6 + 8 + 10 = $
5
$ × $6
$ = $30
$ 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = $
8
$ × $9
$ = $72
答案
③$3$,$4$;④$4$,$5$;$5$,$6$,$30$;$8$,$9$,$72$
解析
① $2=1×2=2$
$n=1$时,等式右边为$1× (1 + 1)$
②$2 + 4=6=2×3$
$n = 2$时,等式右边为$2×(2 + 1)$
通过观察可得规律:从$2$开始$n$个连续偶数相加的和等于$n×(n + 1)$。
③$2+4 + 6$是$3$个连续偶数相加,$n = 3$,所以$2+4+6=3×4$
④$2+4+6 + 8$是$4$个连续偶数相加,$n = 4$,所以$2+4+6+8=4×5$
$2+4+6+8+10$是$5$个连续偶数相加,$n = 5$,所以$2+4+6+8+10=5×6 = 30$
$2+4+6+8+10+12+14+16$是$8$个连续偶数相加,$n = 8$,所以$2+4+6+8+10+12+14+16=8×9=72$
$n=1$时,等式右边为$1× (1 + 1)$
②$2 + 4=6=2×3$
$n = 2$时,等式右边为$2×(2 + 1)$
通过观察可得规律:从$2$开始$n$个连续偶数相加的和等于$n×(n + 1)$。
③$2+4 + 6$是$3$个连续偶数相加,$n = 3$,所以$2+4+6=3×4$
④$2+4+6 + 8$是$4$个连续偶数相加,$n = 4$,所以$2+4+6+8=4×5$
$2+4+6+8+10$是$5$个连续偶数相加,$n = 5$,所以$2+4+6+8+10=5×6 = 30$
$2+4+6+8+10+12+14+16$是$8$个连续偶数相加,$n = 8$,所以$2+4+6+8+10+12+14+16=8×9=72$
2. 看图算一算,填一填。
$ 1 - \frac{1}{2} = \frac{1}{2} $

$ 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} $
$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} $
$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} = $(
$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} = $(
…所以$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - … - \frac{1}{256} = $(
$ 1 - \frac{1}{2} = \frac{1}{2} $
$ 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} $
$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} $
$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} = $(
$\frac{1}{8}$
)$ - $($\frac{1}{16}$
)$ = $($\frac{1}{16}$
)$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} = $(
$\frac{1}{16}$
)$ - $($\frac{1}{32}$
)$ = $($\frac{1}{32}$
)…所以$ 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - … - \frac{1}{256} = $(
$\frac{1}{256}$
)答案
$\frac{1}{8}$,$\frac{1}{16}$,$\frac{1}{16}$;$\frac{1}{16}$,$\frac{1}{32}$,$\frac{1}{32}$;$\frac{1}{256}$
解析
观察前三个算式可知,每次减去的分数是前一个分数的一半,结果等于最后减去的那个分数。
$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} = \frac{1}{8} - \frac{1}{16} = \frac{1}{16}$
$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} = \frac{1}{16} - \frac{1}{32} = \frac{1}{32}$
依此规律,$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - … - \frac{1}{256} = \frac{1}{256}$
$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} = \frac{1}{8} - \frac{1}{16} = \frac{1}{16}$
$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} = \frac{1}{16} - \frac{1}{32} = \frac{1}{32}$
依此规律,$1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - … - \frac{1}{256} = \frac{1}{256}$
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