Prove that `.^(n)C_(0) + .^(n)C_(5) + .^(n)C_(10) + "….."`
`= (2^(n))/(5) (1+2cos^(n)'(pi)/(5)cos'(npi)/(5)+2cos'(pi)/(5)cos'(2npi)/(5))`.

Explanation

Here jump in the series is `'5'`. So, we use fifth roots of unity.
`(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2)x^(2)+.^(n)C_(3)x^(3)+"..."+.^(n)Cx^(n)`
Now, we put `x = 1, alpha, alpha^(2), alpha^(3), alpha^(4)`
where `alpha = cos 'a(2pi)/(5)+isin'(2pi)/(5)`.
Putting these values and then adding, we get
`(1+1)^(n) + (1+alpha)^(n) + (1+alpha^(2))^(n) + (1+alpha^(3))^(n) +(1+alpha^(4))^(n)`
`= 5(.^(n)C_(0) + .^(n)C_(10) + "....")`
`:. 5(.^(n)C_(0)+.^(n)C_(5) +.^(n)C_(10)+".....")`
`= 2^(n)+(1+alpha)^(n)+(1+bar(alpha))^(n)+(1+alpha^(2))^(n)+(1+bar(alpha^(2)))^(n)`
`=2^(n)+2Re(1+cos'(2pi)/(5)+isin'(2pi)/(5))^(n)+2Re(1+cos'(4pi)/(5)+isin'(4pi)/(5))^(n)`
`= 2^(n) + 2Re(2cos^(2) '(pi)/(5)i2sin'(pi)/(5)cos'(pi)/(5))^(n) + 2Re(2cos^(2)'(2pi)/(5)+i2sin'(2pi)/(5)cos'(2pi)/(5))^(n)`
`=2^(n)+2xx2^(n)cos^(n)'(pi)/(5)cos'(npi)/(5)+2xx2^(n)cos^(n)'(2pi)/(5) cos'(2npi)/(5)`
`:. (.^(n)C_(0)+.^(n)C_(5)+.^(n)C_(10)+"......")`
` = (2^(n))/(5)(1+2cos^(n)'(pi)/(5)cos'(npi)/(5)+2cos^(n)'(2pi)/(5)cos'(2npi)/(5))`