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Sequence and Series

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Question : 10 of 86

Marks: +1, -0

The sum of the following is ________.

1+3-5+7+9-11+13+15-17 . . . . .

terms

$2 n^{2}+3$
$5 n^{2}+2$
$3 n^{2}-4 n$
$3 n^{2}$

Solution:

Concept:

The series has a repeating pattern of signs: +, +, –. Group every three consecutive terms to form an arithmetic progression of group sums.

Explanation:

The series is

1+3-5+7+9-11+13+15-17+\ldots

up to

3n

terms.

Group as

(1+3-5)

,

(7+9-11)

,

(13+15-17)

, …

Each group sum:

-1

,

5

,

11

, … which is an A.P. with first term

a=-1

and common difference

d=6

.

Sum of

n

groups:

S_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}[-2+6(n-1)] = \frac{n}{2}[6n-8] = 3n^2-4n

.

Answer:

3n^{2}-4n

(Option C)

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