1/1x3=1/2(1-1/3),1/3x5=1/2(1/3-1/5),1/5x7=1/2(1/5-1/7),……,1/
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1/1x3=1/2(1-1/3),1/3x5=1/2(1/3-1/5),1/5x7=1/2(1/5-1/7),……,1/17x19=1/2(1/17-1/19).∴1/1x3+1/3x5+1/5x7+……+1/17x19
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+……+1/2(1/17-1/19)
=1/2 (1-1/3+1/3-1/5+1/5-1/7+……+1/17-1/19)
=1/2(1-1/19)
=9/19
根据以上材料,回答下列问题:
(1)在式子1/1x3+1/3x5+1/5x7+……中第五项是________
(2)写出第n个 等式,并证明
(3)当m=4时计算1/m(m+4)=1/(m+4)(m+8)+.+1/(m+92)(m+96)的值
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+……+1/2(1/17-1/19)
=1/2 (1-1/3+1/3-1/5+1/5-1/7+……+1/17-1/19)
=1/2(1-1/19)
=9/19
根据以上材料,回答下列问题:
(1)在式子1/1x3+1/3x5+1/5x7+……中第五项是________
(2)写出第n个 等式,并证明
(3)当m=4时计算1/m(m+4)=1/(m+4)(m+8)+.+1/(m+92)(m+96)的值
你写的实在有点乱,应注意用括号!
(1)在式子1/(1x3)+1/(3x5)+1/(5x7)+……中第五项是1/(9×11);
(2)1/(1x3)+1/(3x5)+1/(5x7)+……+1/[(2n-1)(2n+1)]
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+……+1/2[1/(2n-1)-1/(2n+1)]
=1/2 [1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)
(3)1/m(m+4)+1/(m+4)(m+8)+.+1/(m+92)(m+96)
=1/4【1/m-1/(m+4)】+1/4【1/(m+4)-1/(m+8)】+.+1/4【1/(m+92)-1/(m+96)】
=1/4【1/m-1/(m+4)+1/(m+4)-1/(m+8)+.+1/(m+92)-1/(m+96)】
=1/4【1/m-1/(m+96)】
当m=4时,原式=1/4(1/4-1/100)=1/4×24/100=3/50
(1)在式子1/(1x3)+1/(3x5)+1/(5x7)+……中第五项是1/(9×11);
(2)1/(1x3)+1/(3x5)+1/(5x7)+……+1/[(2n-1)(2n+1)]
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+……+1/2[1/(2n-1)-1/(2n+1)]
=1/2 [1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)
(3)1/m(m+4)+1/(m+4)(m+8)+.+1/(m+92)(m+96)
=1/4【1/m-1/(m+4)】+1/4【1/(m+4)-1/(m+8)】+.+1/4【1/(m+92)-1/(m+96)】
=1/4【1/m-1/(m+4)+1/(m+4)-1/(m+8)+.+1/(m+92)-1/(m+96)】
=1/4【1/m-1/(m+96)】
当m=4时,原式=1/4(1/4-1/100)=1/4×24/100=3/50
2/1x3+2/3X5+2/5x7=2/7x9+2/9x11
1/1x3+1/3x5+1/5x7+1/7x9+1/9x11=?
1/1X3+1/3X5+1/5X7...1/17X19+1/19X21=
1/1x3+1/3x5+1/5x7+.+1/2009x2011=
2/1X3+2/3X5+2/5x7+.+2/19x21
(2X2)/1X3+(4X4)/(3X5)+(6X6)/(5X7)+...+(20X20)/(19X21)==?
1/1x3+1/3x5+1/5x7+……+1/99x101
1/1x3+1/3x5+1/5x7+…+1/2011x2013是多少?
已知1-1/3=2/1x3,计算1/1x3+1/3x5+1/5x7+1/7x9+1/9x11=
1X3分之一+3X5分之一+5X7分之一…+2011X2013分之一
C语言编程s(x)=x-x3/3!+x5/5!-x7/7!+……+(-1)n-1·x2n-1/(2n-1)!
sin(x)=x-x3/3!+x5/5!-x7/7!+……+(-1)nx2n+1/(2n+1)!+…… 编程并计算sin