Prove that
`""^(n)C_(3)+""^(n)C_(7) + ""^(n)C_(11) + ...= 1/2{2^(n-1) - 2^(n//2 )sin"" (npi)/(4)}`

Explanation

In given series difference in lower suffices is 4.
i.e., ` 7 -3 =11 = …= 4 `
Now , ` (1)^(1//4) = (cos 0 + I sin 0 )^(1//4)`
`= (cos 2r pi + i sin 2r pi)^(1//4)`
`= cos ""(rpi)/(2) + i sin"" (rpi)/(4) "where r" = 0,1,2,3 `
Four roots of unity ` = 1,i,-1, 1,alpha, alpha^(2) , alpha^(3)` [say]
and ` (1 + x)^(n) = sum_(r=0)^(n) ""^(n)C_(r) x^(r)`
Putting ` x = 1, alpha, alpha^(2) , alpha^(3) , " we get " 2^(n) = sum_(r=0)^(n) ""^(n)C_(r)` ...(i)
` (1 + alpha)^(n) = sum_(r=0)^(n) ""^(n)C_(r) alpha^(r)` ...(ii)
` (1 + alpha ^(2))^(n) = sum_(r=0)^(n) ""^(n)C_(r) alpha^(2r) `...(iii)
`(1 + alpha^(2))^(n) - sum_(r=0)^(n) ""^(n)C_(r) alpha^(3r)` ...(iv)
On multiplying Eq.(i) by 1, Eq.(ii) By ` alpha ` , Eq.(iii) by ` alpha^(2)` and
Eq.(iv) by ` alpha^(3)` and adding , we get
` rArr 2^(n) + alpha (1 + alpha)^(n) + alpha^(2) (1 + alpha^(2))^(n) + alpha^(3) (1 + alpha^(3))^(n)`
` =sum_(r=0)^(n) ""^(n)C_(r) (1 + alpha ^(r+1) + alpha^(2r +2)+ alpha^(3r +3))` ...(v)
for r = 3,7,11,... RHS of Eq.(v)
` =""^(n)C_(r) (1 + alpha ^(4) + alpha^(8)+ alpha^(12))+""^(n)C_(7) (1 + alpha ^(4) + alpha^(16)+ alpha^(24))+""^(n)C_(11) (+ alpha ^(12) + alpha^(24)+ alpha^(36))+ ... `
` = 4(""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11) +...) " " [ because alpha^(4) = 1]`
and LHS of Eq.(v)
`= 2^(n) + ii (1 + i)^(n) + i^(2) (1 + i^(2))^(n) + i^(3) (1 + i^(3))^(n)`
` = 2^(n) + i (1+ i)^(n) + 0 - i (1 - i)^(n)`
` = 2^(n) + i {(1 + i)^(n) - (1 - i)^(n)}`
Since , `[(1 + i)^(n)= [sqrt(2)((1)/(sqrt(2))+ (i)/(sqrt(2)))]^(n)]`
` = 2^(n) + i2^(n//2)* sin"" (npi)/(4) = 2^(n//2) {cos""(pi)/(4) + i sin"" (pi)/(4)}^(n)`
` = 2^(n) - i2^(n//2)* sin"" (npi)/(4) = 2^(n//2) {cos""(pi)/(4) + i sin"" (pi)/(4)}`
Hence , ` 4 (""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11)+ ...) = 2 (2^(n-1) - 2^(n//2) sin" " (npi)/(4))`
` rArr ""^(n)C_(3) + ""^(n)C_(7) + ""^(n)C_(11) + ... = 1/2 (2^(n-1) - 2^(n//2) sin"" (npi)/(4))` .