数列一
1.一个正整数数
如下(表中下一行中数的个数是上一行中数的个数的2倍):
第1行
1
第2行
2 3
第3行
4 5 6 7
…
…
则第9行中的第4个数是( C )
A.132
B.255 C.259
D.260
解析:由数表知表中各行数的个数构成一个以1为首项,公比为2的等比数列.前8行数的个数共有=255(个),故第9行中的第4个数是259.
2.在数列{an}中,a1=1,anan-1=an-1+(-1)n(n≥2, ∈N*),则的值是 ( C )
A. B. C. D.
解析:由已知得a2=1+(-1)2=2,∴a3·a2=a2+(-1)3,∴a3=,∴a4=+(-1)4,∴a4=3,∴3a5=3+(-1)5,∴a5=,∴==.
3.已知数列{an}的前n项和Sn=n3,则a6+a7+a8+a9等于( C )
A.729 B.367 C.604 D.854
解析:a6+a7+a8+a9=S9-S5=93-53=604.
4.已知数列{an}的前n项和Sn=n2-9n,第k项满足5
证明:
解:(1)∵
∴
.
⑵证明:由已知
有
18.已知数列{an}中,a1=0,an+1=an+2n-1(n∈N*).求数列{an}的通项公式an.
解:法一:(累加法)
∵an+1=an+2n-1,
∴an-an-1=2(n-1)-1,
an-1-an-2=2(n-2)-1,
…
a3-a2=2×2-1,
a2-a1=2×1-1.
以上各式左右两边分别相加得
an-a1=2 [1+2+3+…+(n-1)]-(n-1)
=n(n-1)-(n-1)=(n-1)2.
∴an=(n-1)2.
法二:(迭代法)
∵an+1=an+2n-1,
∴an=an-an-1+an-1
=(an-an-1)+(an-1-an-2)+an-2
=…
=(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)+a1
=2(n-1)-1+2(n-2)-1+…+2×2-1+2×1-1+0
=(n-1)2.
19.已知数列
的前
项和
,分别求其通项公式.
⑴
EMBED Equation.3
⑵
EMBED Equation.3
解析:⑴当
,
当
EMBED Equation.3
又
不适合上式,故
②
所以
所以
又
,可知
为等差数列,公差为4
所以
也适合上式,故
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