Q. 224.0( 7 Votes )

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

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} +(–1^{2})^{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}

= (–1^{2}) ^{n}

= 1^{n}

= 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)

= (–1^{2})^{n} (–1)^{1}

= 1^{n} (–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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