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Sequence - Summation and General term

For COMPETITION
Number of Total Problems: 5.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: Understanding

Section:Sequence

Theme:Manipulation
Adjustment# :

Difficulty: 2

Category Summation and General term

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: Understanding

Section:Sequence

Theme:None
Adjustment# :

Difficulty: 2

Category Summation and General term

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: None

Section:Sequence

Theme:None
Adjustment# :

Difficulty: 1

Category Summation and General term

Analysis

Solution/Answer

Problem Num : 4

From : NCTM

Type: Understanding

Section:Sequence

Theme:Pattern
Adjustment# :

Difficulty: 3

Category Summation and General term

Analysis

Solution/Answer

Problem Num : 5

From : AMC10B

Type:

Section:Sequence

Theme:
Adjustment# : 0

Difficulty: 1

'
For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the 2008th term of the sequence?
$mathrm{(A)} {{{2008}}} qquad mathrm{(B)} {{{4015}}} qquad mathrm{(C)} {{{4016}}} qquad mathrm{(D)} {{{4,030,056}}}...$
'

Category Summation and General term

Analysis

Solution/Answer
Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the 2008th term is $2008^2-2007^2=(2008+2007)(2008-2007)=4015Rightarrow oxed{B}$
Note that $n^2$ is the sum of the first n odd numbers.

Answer:

Array ( [0] => 3705 [1] => 3711 [2] => 3739 [3] => 3755 [4] => 8095 ) 5