# Find the value (–1)n + (–1)2n + (–1)2n+ 1 +(–1)4n + 2 where n is any positive odd integer.

(–1)n + (–1)2n + (–1)2n+ 1 +(–1)4n + 2

= (–1)n + (–1)2n + (–1)2n (–1) + (–1)2(2n + 1)

= (–1)n + (–1)2n – (–1)2n +(–12)2n + 1

= –1 + 1 – 1 +1

= 0

Because given, n is a positive odd integer. So (–1)n = –1

2n is a positive EVEN integer

(–1) 2n

= (–12) n

= 1n

= 1

And 2n+1 is a positive ODD integer

(–1) 2n+1 = (–1) 2n (–1) 1

(as per the law of exponents if the bases are same the powers are added)

= (–12)n (–1)1

= 1n (–1)

= –1 ……eq 1

Also, 4n+2 is always a positive even integer.

(–1) 4n+2 = (–1)2(2n + 1) = 1 2n+1 = 1 (from eq 1)

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz:Euclid's Division Lemma44 mins
Fundamental Theorem of Arithmetic-238 mins
NCERT | Imp. Qs. on Rational and Irrational Numbers44 mins
Fundamental Theorem of Arithmetic- 143 mins
Euclids Division Lemma49 mins
Champ Quiz | Fundamental Principle Of Arithmetic41 mins
Application of Euclids Division Lemma50 mins
Relation Between LCM , HCF and Numbers46 mins
Quiz | Fun with Fundamental Theorem of Arithmetic51 mins
NCERT | Fundamental Theorem Of Arithmetic45 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses