1樓:匿名使用者
1/2[2/(1*3)+2/(3*5+2/(5*7)+......+2/(49*51))]
=1/2[1/1-1/3+1/3-1/5+1/5-1/7+......+1/49-1/50]
=1/2[1-1/50]=(1/2)*(49/50)=49/100思路說明:觀察每項的分母的兩個因數(1、3;3、5;5、7;......;49、50)之間差都是2,如果折解為兩個因數的倒數的因式差,則為1/1-1/3=2/1*3; 1/3-1/5=2/3*5; 1/5-1/7=2/5*7;....
;1/49-1/50=2/49*50, 這樣整個序列算式的就無形中擴大了2倍,所以要在拆解式的前面再乘個二分之一(1/2)
2樓:民辦教師小小草
1/(1*3)+1/(3*5)+1/(5*7)+.......+1/(49*51)
=(1-1/3+1/3-1/5+1/5-1/7+.........+1/49-1/51)/2
=(1-1/51)/2
=25/51
1/(1*3)+1/(3*5)+1/(5*7)+....+1/(49*51)=?
3樓:鑑之
原式=1/2(1-1/3+1/3-1/5+1/5-1/7+....+1/49-1/51)
=1/2(1-1/51)
=25/51
1/(1*3)+1/(3*5)+1/(5*7)+......+1/[(2n+1)*(2n+3)]=?
4樓:匿名使用者
1/(1*3)+1/(3*5)+1/(5*7)+......+1/[(2n+1)*(2n+3)]
=(1-1/3+1/3-1/5+1/5-1/7+......+1/(2n+1)-1/(2n+3)*1/2
=(1-1/(2n+3)*1/2
=(2n+2)/(2n+3)*1/2
=(n+1)/(2n+3)
x/(1*2)+x/(2*3)+x/(3*4)+......+x/(n+1)*(n+2)
=x-1/2x+1/2x-1/3x+1/3x-1/4x+......+1/(n+1)x-1/(n+2)x
=x-1/(n+2)x
=(n+1)/(n+2)x
5樓:太白謫仙
1/(1*3)+1/(3*5)+1/(5*7)+......+1/[(2n+1)*(2n+3)]
=1/2(1-1/3+1/3-1/5+...+1/(2n+1)-1/(2n+3))
=1/2(1-1/(2n+3))
=(n+1)/(2n+3)
x/(1*2)+x/(2*3)+x/(3*4)+......+x/(n+1)*(n+2)
=x(1-1/2+1/2-1/3+...+1/(n+1)-1/(n+2))
=x(1-1/(n+2))
=x(n+1)/(n+2)
6樓:我不是他舅
1/(1*3)+1/(3*5)+1/(5*7)+......+1/[(2n+1)*(2n+3)]
=/2=/2
=[1-1/(2n+3)]/2
=(n+1)/(2n+3)
x/(1*2)+x/(2*3)+x/(3*4)+......+x/(n+1)*(n+2)
=x*[1/(1*2)+1/(2*3)+1/(3*4)+......+1/(n+1)*(n+2)]
=x*[(1-1/2)+(1/2-1/3)+……+[1/(n+1)-1/(n+2)]
=x*[1-1/(n+2)]
=(n+1)x/(n+2)
7樓:匿名使用者
因為:1/[(2n+1)*(2n+3)]==(1/(2n+1)-1/(2n+3))/2;
所以:1/(1*3)+1/(3*5)+1/(5*7)+......+1/[(2n+1)*(2n+3)]
=[1-1/3+1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)]/2
=[1-1/(2n+3)]/2
=(n+1)/(2n+3)
8樓:
根據裂項求和來做!只是由於分母不是相鄰的所以要誠意它們的差值分之一而已!
9樓:洪雁楣
這是一道程式設計題目,不會,問老師!
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