3.单项选择

1)     Lisa! I’m sorry --- I _______ recognize you --- you ________ your hair cut!

A. don’t; had    B. didn’t; have had  C. don’t; have had  D. didn’t had

2)     The recording was __________. I can know exactly what happened at that time.

A. bad quality   B. good quality   C. of bad quality  D. of good quality

3)     Buy vegetables _______ small quantities for your immediate use.

A. in   B. with   C. at   D. on

4)     Tanya, be __________! Our baby is sleeping and you may wake him up.

A. silent   B. still   C. quiet   D. noisy

5)     It remains ________ whether the operation was successful.

A. to be seen   B. to see   C. seeing   D. to have seen

6)     Australia’s unemployment ______ rose to 6.5% in February.

A. percent   B. rank   C. level   D. rate

7)     --- Can you see the sign on the door? --- Yes, it ________ “No Entry”.

A. reads   B. reflects   C. registers  D. reports

8)     We regret _______ you that your application has not been successful.

A. informing   B. to inform  C being informed   D. to have informed

9)     When the head teacher said someone didn’t obey the class discipline, we all knew whom he was ____________.

A. replying to   B. reacting to   C. referring to   D. relating to

10)   The terrible situation required that the manager ________ present.

A. be    B. was   C. would be  D. shall be

解法二  设Sx=Ax2+Bx(x∈N)

①-②,得A(m2-n2)+B(m-n)=n-m

∵m≠n  ∴ A(m+n)+B=-1

故A(m+n)2+B(m+n)=-(m+n)

即Sm+n=-(m+n)

说明  a1,d是等差数列的基本元素,通常是先求出基本元素,再

解的“整体化”思想,在解有关数列题目中值得借鉴.解法二中,由于是等差数列,由例22,故可设Sx=Ax2+Bx.(x∈N)

[例14]  在项数为2n的等差数列中,各奇数项之和为75,各偶数项之和为90,末项与首项之差为27,则n之值是多少?

解  ∵S偶项-S奇项=nd

∴nd=90-75=15

又由a2n-a1=27,即(2n-1)d=27

[例15]  在等差数列{an}中,已知a1=25,S9=S17,问数列前多少项和最大,并求出最大值.

解法一  建立Sn关于n的函数,运用函数思想,求最大值.

∵a1=25,S17=S9  解得d=-2

∴当n=13时,Sn最大,最大值S13=169

解法二  因为a1=25>0,d=-2<0,所以数列{an}是递减等

∵a1=25,S9=S17

∴an=25+(n-1)(-2)=-2n+27

即前13项和最大,由等差数列的前n项和公式可求得S13=169.

解法三  利用S9=S17寻找相邻项的关系.

由题意S9=S17得a10+a11+a12+…+a17=0

而a10+a17=a11+a16=a12+a15=a13+a14

∴a13+a14=0,a13=-a14  ∴a13≥0,a14≤0

∴S13=169最大.

解法四  根据等差数列前n项和的函数图像,确定取最大值时的n.

∵{an}是等差数列

∴可设Sn=An2+Bn

二次函数y=Ax2+Bx的图像过原点,如图3.2-1所示

∵S9=S17

∴取n=13时,S13=169最大

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