Question

(0.1²+0.2²+0.3²+0.4²)²/(0.1³+0.2³+0.3³+0.4³)³

简算！有根据！

Answer 1

解： (0.1²+0.2²+0.3²+0.4²)²/(0.1³+0.2³+0.3³+0.4³)³
=[0.01（1+4+9+16） ]² / [0.001(1+8+27+64)]³
=0.3² /0.1³
=90
解法二：1^1+2^2+……+n^2=1/6*n(2n+1)(n+1)
1^3+2^3+3^3+……+n^3=(1+2+3+……+n)^2
所以
(0.1²+0.2²+0.3²+0.4²)²/(0.1³+0.2³+0.3³+0.4³)³
= (0.1²×1³+0.1²×2²+0.1²×3²+0.1²×4²)²/(0.1³×1³+0.1³×2³+0.1³×3³+0.1³×4³)³
=0.01²(1³+2²+3²+4²)²/[0.001³(1³+2³+3³+4³)³]
=0.01²·[1/6·4(4+1)（8+1）]² / {0.001³×[(1+2+3+4)²]³}
=0.01²·[30]² / {0.001³×[(100]³}
=0.3² /0.1³
=90

Answer 2

解：根据：1^1+2^2+……+n^2=1/6*n(2n+1)(n+1)
1^3+2^3+3^3+……+n^3=(1+2+3+……+n)^2 可以简算：
(0.1²+0.2²+0.3²+0.4²)²/(0.1³+0.2³+0.3³+0.4³)³
=1/6*0.4(0.4+1)(0.8+1)/1^2
=0.168 .