数学题奥数题：1⼀（1×2×3×4）+3⼀（2×3×4×5）+5⼀（3×4×5×6）+7⼀（4×5×6×7）+……

通式 ∑(2n-1)/[n(n+1)(n+2)(n+3)]
化为 ∑{1/[(n(n+2)]-1/[(n+1)(n+3)]-2/[n(n+3)]+2/[(n+1)(n+2)]}
分别求解
∑1/[(n(n+2)] = 1/2-1/2(n+2)
∑1/[(n+1)(n+3)] = 1/4 - 1/2(n+3)
∑2/[n(n+3)] = 2/3 - 2/3(n+3)
∑2/[(n+1)(n+2)] = 1-2/(n+2)
综合，
∑(2n-1)/[n(n+1)(n+2)(n+3)]
= [1/2-1/2(n+2)] - [1/4 - 1/2(n+3)] - [2/3 - 2/3(n+3)] + [1-2/(n+2)]
= 7/12 - 5/2(n+2) + 7/6(n+3)
取 n=9
1/（1×2×3×4）+3/（2×3×4×5）+5/（3×4×5×6）+7/（4×5×6×7）+9/（5×6×7×8）+……+17/（9×10×11×12）
= 7/12 - 5/22 + 7/72
= 359/792

交你个简单的算法，先用计算器将括号内的得数算出来，在把得到每一个分数相加，最后就可以以分数加法的算法计算了

第一步,写通式
为 (2n-1)/n(n+1)(n+2)(n+3)
第二步，在通式因子找到裂项相消的项。 肯定是1/n(n+i)-1/(n+?)(n+?)
分别把i=1,2,3代入。发现得到
i=1:
1/n(n+1)-1/(n+2)(n+3)=(4n+6)/n(n+1)(n+2)(n+3) 2a+b=1 6a+4b=1
i=2:
1/n(n+2)-1/(n+1)(n+3)=(2n-4)/n(n+1)(n+2)(n+3) 6a+3b=3 6a-4b=-1
i=3:
...=2/n(n+1)(n+2)(n+3)

所以 把分子2n-1分成 (2n-4)+3/2 *2
就得到原式=∑(2n-1)/[n(n+1)(n+2)(n+3)]
=5/14∑(4n-6)/[n(n+1)(n+2)(n+3)]+2/7∑(2n-4)/[n(n+1)(n+2)(n+3)]
=5/14∑(1/n(n+1)-1/(n+2)(n+3))+2/7∑(1/n(n+2)-1/(n+1)(n+3))
=5/14*(1/(1*2)+1/(2*3)-1/(10*11)-1/(11*12))-2/7*(1/(1*3)-1/(10*12))
计算得到=39/280

不知道

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