数列:an+1(n+1为下标)+an=3n-54,若Sn为前n项和,求证:当a1>-27时,存在n,使Sn最小
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数列:an+1(n+1为下标)+an=3n-54,若Sn为前n项和,求证:当a1>-27时,存在n,使Sn最小
当n=2k,k为正整数时,
Sn=(a1+a2)+(a3+a4)+(a5+a6)+-----+(a(n-1)+an)
=(3-54)+(3*3-54)+(3*5-54)+----+(3(2k-1)-54)
=3[1+3+5+----+(2k-1)]-54k
=3k^2-54k
=(3/4)n^2-27n
当n=2k+1,k为正整数时
Sn=a1+(a2+a3)+(a4+a5)+-----(a(2k)+a(2k+1))
=a1+(3*2-54)+(3*4-54)+----(3*2k-54)
=a1+3k^2-51k
=a1+(3/4)n^2-27n+(105/4)
由上知Sn为关于n的二次函数,且二次项系数为正,所以Sn必有最小值.
Sn=(a1+a2)+(a3+a4)+(a5+a6)+-----+(a(n-1)+an)
=(3-54)+(3*3-54)+(3*5-54)+----+(3(2k-1)-54)
=3[1+3+5+----+(2k-1)]-54k
=3k^2-54k
=(3/4)n^2-27n
当n=2k+1,k为正整数时
Sn=a1+(a2+a3)+(a4+a5)+-----(a(2k)+a(2k+1))
=a1+(3*2-54)+(3*4-54)+----(3*2k-54)
=a1+3k^2-51k
=a1+(3/4)n^2-27n+(105/4)
由上知Sn为关于n的二次函数,且二次项系数为正,所以Sn必有最小值.
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