初中数学............
1个回答
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(1)观察并猜想:
12+22=(1+0)×1+(1+1)×2=l+0×1+2+1×2=(1+2)+(0×1+1×2)
12+22+32=(1+0)×1+(1+1)×2+(l+2)×3
=1+0×1+2+1×2+3+2×3
=(1+2+3)+(0×1+1×2+2×3)
12+22+32+42=(1+0)×1+(1+1)×2+(l+2)×3+
(1+3)×4
=1+0×1+2+1×2+3+2×3+
4+3×4
=(1+2+3+4)+(
0×1+1×2+2×3+3×4
)
…
(2)归纳结论:
12+22+32+…+n2=(1+0)×1+(1+1)×2+(1+2)×3+…[1+(n-l)]n
=1+0×1+2+1×2+3+2×3+…+n+(n-1)×n
=(
1+2+3+…+n
)+[
0×1+1×2+2×3+…+(n-1)n
]
=
n(n+1)
+
n(n+1)(n-1)
=×
n(n+1)(2n+1)
12+22=(1+0)×1+(1+1)×2=l+0×1+2+1×2=(1+2)+(0×1+1×2)
12+22+32=(1+0)×1+(1+1)×2+(l+2)×3
=1+0×1+2+1×2+3+2×3
=(1+2+3)+(0×1+1×2+2×3)
12+22+32+42=(1+0)×1+(1+1)×2+(l+2)×3+
(1+3)×4
=1+0×1+2+1×2+3+2×3+
4+3×4
=(1+2+3+4)+(
0×1+1×2+2×3+3×4
)
…
(2)归纳结论:
12+22+32+…+n2=(1+0)×1+(1+1)×2+(1+2)×3+…[1+(n-l)]n
=1+0×1+2+1×2+3+2×3+…+n+(n-1)×n
=(
1+2+3+…+n
)+[
0×1+1×2+2×3+…+(n-1)n
]
=
n(n+1)
+
n(n+1)(n-1)
=×
n(n+1)(2n+1)
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