Given that n is a positive integer.
We have to prove
(3+1)n+(3−1)n=2n+1cosπn6( \sqrt{3}+1 )^n +( \sqrt{3}-1)^n= 2^{n+1} cos\frac{\pi n }{6}(3+1)n+(3−1)n=2n+1cos6πn
Question is wrong
Take n=2n=2n=2 .
L.H.S=(3+1)2+(3−1)2L.H.S=(\sqrt{3}+1)^2+(\sqrt {3} -1)^2L.H.S=(3+1)2+(3−1)2
=3+1+23+3+1−23=3+1+2\sqrt{3}+3+1-2\sqrt {3}=3+1+23+3+1−23
=8=8=8
R.H.S=22+1cos2π6R.H.S=2^{2+1} cos\frac{ 2\pi }{6}R.H.S=22+1cos62π =23×12=4=2^3×\frac{1}{2}=4=23×21=4
Hence , L.H.S≠R.H.SL.H.S \neq R.H.SL.H.S=R.H.S
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