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 +(–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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