CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

The series ...

Question

The series $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . + 99^{2} - 100^{2} =$ _____

A

- 5050

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

5050

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

11000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

- 11000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 5050$
The sum of nth term of the series is given as
$S = (- 1)^{n + 1} \frac{n (n + 1)}{2}$
Here $n = 100$
So,
$S = (- 1)^{100 + 1} \frac{100 (100 + 1)}{2}$
$= (- 1) \frac{10100}{2}$
$= - 5050$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The value of

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . + 99^{2} - 100^{2}

Q. The sum of the series

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . - 100^{2} ?

Q. The value of

100^{2} - 99^{2} + 98^{2} - 97^{2} . . . . . . . . . - 3^{2} + 2^{2} - 1^{2}

Q. Observe the following pattern
2² − 1² = 2 + 1
3² − 2² = 3 + 2
4² − 3² = 4 + 3
5² − 4² = 5 + 4
and find the value of
(i) 100² − 99²
(ii) 111² − 109²
(iii) 99² − 96²

Q. The least value of

(x + 100)^{2} + (x + 99)^{2} + \dots \dots + (x + 1)^{2} + x^{2} + (x - 1)^{2} + (x - 2)^{2} + \dots . + (x - 100)^{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Summation by Sigma Method

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Summation by Sigma Method

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon