一个数列的运算1/2x4+1/4x6+1/6x8+……+1/2k(2k+2)+1/(2k+2)(2k+4)连续偶数的乘积相加有特定的算法还是怎么?我看答案上面写的就直接=k/4(k+1)+1/4(k+1)(k+2)答案那样是因为后面需要这样的运算那我按照你
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![一个数列的运算1/2x4+1/4x6+1/6x8+……+1/2k(2k+2)+1/(2k+2)(2k+4)连续偶数的乘积相加有特定的算法还是怎么?我看答案上面写的就直接=k/4(k+1)+1/4(k+1)(k+2)答案那样是因为后面需要这样的运算那我按照你](/uploads/image/z/2718928-64-8.jpg?t=%E4%B8%80%E4%B8%AA%E6%95%B0%E5%88%97%E7%9A%84%E8%BF%90%E7%AE%971%2F2x4%2B1%2F4x6%2B1%2F6x8%2B%E2%80%A6%E2%80%A6%2B1%2F2k%282k%2B2%29%2B1%2F%282k%2B2%29%282k%2B4%29%E8%BF%9E%E7%BB%AD%E5%81%B6%E6%95%B0%E7%9A%84%E4%B9%98%E7%A7%AF%E7%9B%B8%E5%8A%A0%E6%9C%89%E7%89%B9%E5%AE%9A%E7%9A%84%E7%AE%97%E6%B3%95%E8%BF%98%E6%98%AF%E6%80%8E%E4%B9%88%3F%E6%88%91%E7%9C%8B%E7%AD%94%E6%A1%88%E4%B8%8A%E9%9D%A2%E5%86%99%E7%9A%84%E5%B0%B1%E7%9B%B4%E6%8E%A5%3Dk%2F4%28k%2B1%29%2B1%2F4%28k%2B1%29%28k%2B2%29%E7%AD%94%E6%A1%88%E9%82%A3%E6%A0%B7%E6%98%AF%E5%9B%A0%E4%B8%BA%E5%90%8E%E9%9D%A2%E9%9C%80%E8%A6%81%E8%BF%99%E6%A0%B7%E7%9A%84%E8%BF%90%E7%AE%97%E9%82%A3%E6%88%91%E6%8C%89%E7%85%A7%E4%BD%A0)
一个数列的运算1/2x4+1/4x6+1/6x8+……+1/2k(2k+2)+1/(2k+2)(2k+4)连续偶数的乘积相加有特定的算法还是怎么?我看答案上面写的就直接=k/4(k+1)+1/4(k+1)(k+2)答案那样是因为后面需要这样的运算那我按照你
一个数列的运算
1/2x4+1/4x6+1/6x8+……+1/2k(2k+2)+1/(2k+2)(2k+4)
连续偶数的乘积相加有特定的算法还是怎么?
我看答案上面写的就直接
=k/4(k+1)+1/4(k+1)(k+2)
答案那样是因为后面需要这样的运算
那我按照你们给的算法能不能得到我说的答案啊?
一个数列的运算1/2x4+1/4x6+1/6x8+……+1/2k(2k+2)+1/(2k+2)(2k+4)连续偶数的乘积相加有特定的算法还是怎么?我看答案上面写的就直接=k/4(k+1)+1/4(k+1)(k+2)答案那样是因为后面需要这样的运算那我按照你
1/2x4+1/4x6++……+1/2k(2k+2)+1/(2k+2)(2k+4)
=(1/2)x[2/2x4+2/4x6+……+2/2k(2k+2)+2/(2k+2)(2k+4]
=(1/2)x[1/2-1/4+1/4-1/6+……+1/2k-1/(2k+2)+1/(2k+2)-1/(2k+4]
中间正负抵消
=(1/2)x[1/2-1/(2k+4]
=1/4-1/(4k+8)
=(k+1)/(4k+8)
他的这个答案也对,但是没有化到最简
1/2*4=1/2(1/2-1/4)
1/4*6=1/2(1/4-1/6)
依次展开就可以运算了
这个数列通项为
a(n) = 1/(2n+2)(2n+4) = 1/2 * [1/(2n+2) - 1/(2n+4)]
把这数列按 n = 0 到 k 求和,得到
1/2 * [1/2 - 1/4] + 1/2 * [1/4 - 1/6] + 1/2 * [1/6 - 1/8] + ... + 1/2 * [1/(2k+2) - 1/(2k+4)]
相邻两项能抵消,...
全部展开
这个数列通项为
a(n) = 1/(2n+2)(2n+4) = 1/2 * [1/(2n+2) - 1/(2n+4)]
把这数列按 n = 0 到 k 求和,得到
1/2 * [1/2 - 1/4] + 1/2 * [1/4 - 1/6] + 1/2 * [1/6 - 1/8] + ... + 1/2 * [1/(2k+2) - 1/(2k+4)]
相邻两项能抵消,最后就剩下
1/2 * [1/2 - 1/(2k+4)]
= 1/4 - 1/(4k+8)
= (k+2-1) / (4k+8)
= (k+1) / (4k+8)
你那个答案能化简,
k/4(k+1)+1/4(k+1)(k+2)
= k(k+2)+1 / 4(k+1)(k+2)
= (k^2 + 2k + 1) / 4(k+1)(k+2)
= (k+1)^2 / 4(k+1)(k+2)
= (k+1) / (4k+8)
跟我那答案一样。
收起
1/(2k+2)(2k+4)展开得1/4*[1/(k+1)-1/(k+2)]
剩下自己做
1/2x4=(1/2-1/4)/2
1/4x6=(1/4-1/6)/2
1/6x8=(1/6-1/8)/2
.
.
.
1/2k(2k+2)=[1/2k-1/(2k+2)]/2
1/(2k+2)(2k+4)=[1/(2k+2)-1/(2k+4)]/2
上面各式相加,两两+-抵消最后剩下两项(1/2)/2-[1/(2k+4)]/2,即
1/4-1/(4k+8)=1/4-1/4(k+2)=(k+1)/4(k+2),和你的答案并不相同。