題目有誤
是否存在a,b,c,使等式1*2+3*2+5*2+```+(2n-1)*2=1/3an(bn*2+c)對(duì)任意正整數(shù)n都成
存在,a=1,b=4,c=-1.詳解如下.
解 根據(jù)自然數(shù)2方冪和公式得：
1^2+2^2+3^2+……n^2=n(n+1)(2n+1)/6
所以
1^2+2^2+3^2+……(2n)^2=n(2n+1)(4n+1)/3
故有
1^2+3^2+5^2+……(2n-1)^2=n(2n+1)(4n+1)/3-4[n(n+1)(2n+1)/6]
=n(2n-1)(2n+1)/3=n(4n^2-1)/3
所以a=1,b=4,c=-1.
或者根據(jù)通項(xiàng):an=4n^2-4n+1,
Sn=4n(n+1)(2n+1)/6-4n(n+1)/2+n=n(4n^2-1)/3.