Let l(n) = 2cosnx, n belong to natural number then l(1) l(n+1) l(n)

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Solution

f(1) = 2 cosx
f(n+1) = 2 cos(n+1)x
f(n) = 2 cos nx

now

f(1) f(n+1) = 2 [cos (n+2)x + cos nx]

hence

f(1) f(n+1) – f(n) = 2 cos (n+2)x

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