Question

已知數(shù)列{a_n}是首項(xiàng)為1、公差為2的等差數(shù)列,對(duì)每一個(gè)k∈N^*,在a_k與a_k+1之間插入2^k-1個(gè)2,得到新數(shù)列{b_n}.設(shè)S_n、T_n分別是數(shù)列{b_n}和{a_n}的前n項(xiàng)和.

(1)試問(wèn)a₁₀是數(shù)列{b_n}的第幾項(xiàng)?

(2)是否存在正整數(shù)m,使S_m=2 008?若存在,求出m的值;若不存在,請(qǐng)說(shuō)明理由.

(3)若a_m是數(shù)列{b_n}的第f(m)項(xiàng),試比較S_f(m)與2T_m的大小,并說(shuō)明理由.

Answer 1

解:(1)∵在數(shù)列{b_n}中,對(duì)每一個(gè)k∈N,在a_k與a_k+1之間有2^k-1個(gè)2,

∴a₁₀在數(shù)列{b_n}中的項(xiàng)數(shù)為10+1+2+4+…+2⁸==521,

即a₁₀是數(shù)列{b_n}中第521項(xiàng).                                         

(2)a_n=1+(n-1)·2=2n-1,

在數(shù)列{b_n}中,a_m及其前面所有項(xiàng)的和為［1+3+5+…+(2m-1)］+(2+4+…+2^m-1)

=m²+=2^m+m²-2,                                         

∵2¹⁰+10-2=1 122＜2 008＜2¹¹+11-2,且2 008-1 122=886=443×2,

∴存在m=521+443=964,使得S_m=2 008.                                

(3)方法一:由(2)知S_f(m)=2^m+m²-2,

又T_m=1+3+5+…+(2m-1)=m²，

所以S_f(m)-2T_m=(2^m+m²-2)-2m²=2^m-(m²+2),                               

要比較S_f(m)與2T_m的大小,只需比較2^m與m²+2的大小即可.

當(dāng)m=1時(shí),2^m=2,m²+2=3,故2^m＜m²+2;

當(dāng)m=2時(shí),2^m=4,m²+2=6,故2^m＜m²+2;

當(dāng)m=3時(shí),2^m=8,m²+2=11,故2^m＜m²+2;

當(dāng)m=4時(shí),2^m=16,m²+2=18,故2^m＜m²+2;                                

當(dāng)m≥5時(shí),2^m=(1+1)^m=1+C¹_m+C²_m+…+++1

≥1+m+++m+1=m²+m+2＞m²+2.

故當(dāng)m=1,2,3,4時(shí)，S_f(m)＜2T_m;

當(dāng)m≥5且m∈N時(shí),S_f(m)＞2T_m.                                       

方法二:以上同方法一,

當(dāng)m≥5時(shí),∵m-2≥3,

∴2^m=(1+1)^m=1++++…+1

≥1+m+++1

≥1+m+++1=m²+2.

故當(dāng)m=1,2,3,4時(shí),S_f(m)＜2T_m;

當(dāng)m≥5且m∈N時(shí),S_f(m)＞2T_m.                                     

(另法提示:令c_m=2^m-(m²+2),d_m=c_m+1-c_m=2^m-(2m+1),由d_m+1-d_m=2^m-2≥0，可知d_m+1≥d_m，又d₃＞0,故對(duì)任意的m≥3,d_m＞0,從而數(shù)列{c_m}從第3項(xiàng)起往后是遞增的,又c₅=5＞0,故當(dāng)m≥5時(shí),c_m＞0,即S_f(m)＞2T_m).