an=1/[(2n+1)(2n+3)],求Sn.也就是:an等于1除以(2n+1)(2n+3)的商,求Sn.我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/(2n+1)-1/(2n+3)],接下来怎么办?
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![an=1/[(2n+1)(2n+3)],求Sn.也就是:an等于1除以(2n+1)(2n+3)的商,求Sn.我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/(2n+1)-1/(2n+3)],接下来怎么办?](/uploads/image/z/8984581-61-1.jpg?t=an%3D1%2F%5B%282n%2B1%29%282n%2B3%29%5D%2C%E6%B1%82Sn.%E4%B9%9F%E5%B0%B1%E6%98%AF%EF%BC%9Aan%E7%AD%89%E4%BA%8E1%E9%99%A4%E4%BB%A5%EF%BC%882n%2B1%EF%BC%89%282n%2B3%29%E7%9A%84%E5%95%86%2C%E6%B1%82Sn.%E6%88%91%E5%B7%B2%E7%BB%8F%E7%AE%97%E5%88%B0Sn%3D1%2F2%5B1%2F%282%2B1%29+-1%2F%282%2B3%29+%2B+1%2F%284%2B1%29-1%2F%284%2B3%29%E2%80%A6%2B1%2F%EF%BC%882n%2B1%EF%BC%89-1%2F%282n%2B3%29%5D%2C%E6%8E%A5%E4%B8%8B%E6%9D%A5%E6%80%8E%E4%B9%88%E5%8A%9E%3F)
an=1/[(2n+1)(2n+3)],求Sn.也就是:an等于1除以(2n+1)(2n+3)的商,求Sn.我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/(2n+1)-1/(2n+3)],接下来怎么办?
an=1/[(2n+1)(2n+3)],求Sn.
也就是:an等于1除以(2n+1)(2n+3)的商,求Sn.
我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/(2n+1)-1/(2n+3)],接下来怎么办?
an=1/[(2n+1)(2n+3)],求Sn.也就是:an等于1除以(2n+1)(2n+3)的商,求Sn.我已经算到Sn=1/2[1/(2+1) -1/(2+3) + 1/(4+1)-1/(4+3)…+1/(2n+1)-1/(2n+3)],接下来怎么办?
an=1/[(2n+1)(2n+3)]=1/2[1/(2n+1)-1/(2n+3)]
Sn=`1/2[(1/3-1/5)+(1/5-1/7)+.+(1/(2n+1)-1/(2n+3)]
=1/2[1/3-1/(2n+3)]
=1/6-1/(4n+6)
对 就是 这么做的啊
结果变成 1/2倍的 (1/3-1/5+1/5-1/7。。。。+1/(2n+1)-1/(2n+3)
最后为 1/2倍的(1/3-1/(2n+3))
都算到这了,还不会阿,加减可以抵消的,唉,坚持一下就出来了阿
an=[1/(2n+1)-1/(2n+3)]/2
Sn=a1+a2+a3+...........+an
=[(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+......+(1/(2n+1)-1/(2n+3))]/2
=[1/3-1/(2n+3)]/2
=n/(6n+9)
注意1/(....)(......)
可以用裂项相消法
你那样做是对的呀!!
然后接着做就是了
即Sn=1/2[1/3-1/5+1/5-1/7~~~~+1/(2n-1)-1/(2n+1)+1/(2n+1)-1/(2n+3)]
解出Sn=1/2[1/3-1/(2n+3)]=1/6