求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
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求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
![求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn](/uploads/image/z/5306576-32-6.jpg?t=%E6%B1%82%E6%95%B0%E5%88%971%2F2%5E2%2B4%2C1%2F4%5E2%2B8%2C1%2F6%5E2%2B12.1%2F%282n%29%5E2%2B4n%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%92%8Csn)
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
consider
1/[(2n)^2+4n]
=1/[4n(n+1)]
=(1/4)[1/n -1/(n+1)]
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
=(1/4) [ (1/1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)]
=(1/4)( 1-1/(n+1) )
consider
1/[(2n)^2+4n]
=1/[4n(n+1)]
=(1/4)[1/n -1/(n+1)]
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]
=(1/4) [ (1/1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)]
=(1/4)( 1-1/(n+1) )
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