(–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= 0Because given, n is a positive odd integer. So (–1)n = –12n is a positive EVEN integer⇒ (–1) 2n= (–12) n= 1n= 1And 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 1Also, 4n+2 is always a positive even integer.(–1) 4n+2 = (–1)2(2n + 1) = 1 2n+1 = 1 (from eq 1)

Q. 224.0( 7 Votes )

Find the value (–

Class 10th

Answer :

(–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)

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