数列an前N项为Sn满足Sn=n²+n.1.求an通项公式 2.令bn=(n+1)/(n+2)²an² 数列bn前N项和为Tn,证明Tn小于5/64
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![数列an前N项为Sn满足Sn=n²+n.1.求an通项公式 2.令bn=(n+1)/(n+2)²an² 数列bn前N项和为Tn,证明Tn小于5/64](/uploads/image/z/1550130-42-0.jpg?t=%E6%95%B0%E5%88%97an%E5%89%8DN%E9%A1%B9%E4%B8%BASn%E6%BB%A1%E8%B6%B3Sn%3Dn%26%23178%3B%2Bn.1.%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+2.%E4%BB%A4bn%3D%28n%2B1%29%2F%28n%2B2%29%26%23178%3Ban%26%23178%3B+%E6%95%B0%E5%88%97bn%E5%89%8DN%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%AF%81%E6%98%8ETn%E5%B0%8F%E4%BA%8E5%2F64)
数列an前N项为Sn满足Sn=n²+n.1.求an通项公式 2.令bn=(n+1)/(n+2)²an² 数列bn前N项和为Tn,证明Tn小于5/64
数列an前N项为Sn满足Sn=n²+n.1.求an通项公式 2.令bn=(n+1)/(n+2)²an² 数列bn前N项和为Tn,证明Tn小于5/64
数列an前N项为Sn满足Sn=n²+n.1.求an通项公式 2.令bn=(n+1)/(n+2)²an² 数列bn前N项和为Tn,证明Tn小于5/64
Sn=n²+n
(1)n=1时,
a1=S1=1+1=2
(2)n≥2时,
an=Sn-S(n-1)=n²+n-(n-1)²-(n-1)=2n
n=1时也满足
∴ an=2n
二、
bn=(n+1)/[(n+2)²*(2n)²]=[4(n+1)/16]/[(n+2)²*n²]=(1/16)*[1/n² -1/(n+2)²]
∴ Tn=(1/16)*[1/1-1/9+1/4-1/16+.+1/(n-1)²-1/(n+1)²+1/n² -1/(n+2)²]
=(1/16)[1+1/4-1/(n+1)²-1/(n+2)²]