Prove that `cos4x = 2cos^2(2x) - 1`

Explanation

`T_(r ) = T_((r-1)+1) = ""^(n)C_(r-1) x^(r-1)`
` T_(r +1) = ""^(n)C_(r) x^(r) " and " T_(r +2) = T_((r +1)+1)= ""^(n)C_(r +2) x^(r+1)`
` therefore ` Coefficient of rth, (r +1)th and (r +2)th terms in the
expansion of
`(1 + x)^(n) "are" ""^(n)C_(r-1) ""^(n)C_(r) , ""^(n)C_(r+1)`
` because " Given " , ""^(n)C_(r-1) . ""^(n)C_(r) , ""^(n)C_(r+1) ` are in AP
and ` n ge r + 1 `
`(""^(n)C_(r-1))/(""^(n)C_(r)) ,1,(""^(n)C_(r+1) )/(""^(n)C_(r))` are also in AP
` rArr (r)/(n-r +1) = (n-r)/(r+1) - 1 rArr (n-2r +1)/(n-r +1) = (n-2r -)/(r +1)`
` rArr nr - 2r^(2) + r + n - 2r + 1`
`= n^(2) - 2nr - n- nr + 2r^(2) + r + n - 2r - 1`
` rArr n^(2) - 4nr^(2) = n + 2 rArr (n - 2r(^(2) = n +2`
Corollary I for ` r = 2, n = 7 " "[ because n ge 3 ] `
Corollary II for r = 5, n = 7,14 ` " "[ because n ge 6 ] ` .