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計算 [(2^4+1/4)(4^4+1/4)(6^4+1/4)(8^4+1/4)(10^4+1/4)]/[(1^4+1/4)(3^4+1/4)(5^4+1/4)(7^4+1/4)(9^4+1/4)] =?

2005-09-28 06:00:46 · 2 個解答 · 發問者 蔡春益 7 in 科學 數學

2 個解答

因為a4+1/4=(a4+a2+1/4)-a2=(a2+1/2)2-a2=(a2+a+1/2)(a2-a+1/2)=[a(a-1)+1/2][a(a+1)+1/2]原式分子=(1*2+1/2)(2*3+1/2)(3*4+1/2)(4*5+1/2)(5*6+1/2)(6*7+1/2)(7*8+1/2)(8*9+1/2)(9*10+1/2)(10*11+1/2)=Π(k=1到10)[k(k+1)+1/2]原式分母=(0*1+1/2)(1*2+1/2)(2*3+1/2)(3*4+1/2)(4*5+1/2)(5*6+1/2)(6*7+1/2)(7*8+1/2)(8*9+1/2)(9*10+1/2)=Π(k=0到9)[k(k+1)+1/2]原式=(10*11+1/2)/(0*1+1/2)=110.5/0.5=221

2005-09-28 10:11:01 · answer #1 · answered by ? 7 · 0 0

應該是221吧...
>_<
不過這個問題好奇怪喔= =

2005-09-28 06:07:47 · answer #2 · answered by Anonymous · 0 0

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