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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