Find the sum `sum_(r=0)^(5)""^(32)C_(6r)`.

Explanation

Consider `(1+x)^(32) = .^(32)C_(0) + .^(32)C_(1)x + .^(32)C_(2)x^(2) + "….." + .^(32)C_(32)x^(32)`
In `undersetr(r=0)overset(5)sum.^(32)C_(6r)` there is a jump of `'6'` in binomial coefficients, so are will use sixth roots unity.
Puttingl `x = cos'(2pir)/(6) + isin'(2pir)/(6), r = 0, 1,2,3,4,5` and adding .
we get
`6[.^(32)C_(0) + .^(32)C_(6)+.^(32)C_(12)+"....."+.^(32)C_(30)]`
`=(1+1)^(32)+(1-1)^(32)+[(3/2+i'(sqrt(3))/(2))^(32)+(3/2-i'(sqrt(3))/(2))^(32)]+[(1/2+i'(sqrt(3))/(2))^(32)+(1/2-i'(sqrt(3))/(2))^(32)]`
`=2^(32)+(sqrt(3))^(32)[(cos'(pi)/(6)+isin'(pi)/(6))^(32)+(cos'(pi)/(6)-isin'(pi)/(6))^(32)]+[(cos'(pi)/(3)+isin'(pi)/(3))^(32)+(cos'(pi)/(3)-isin'(pi)/(3))^(32)]`
`= 2^(32) + 3^(16)[2cos'(32pi)/(6)]+[2cos'(32pi)/(3)]`
`=2^(32)+3^(16)[2cos(5pi+(pi)/(3))]+[2cos(11pi-pi/3)]`
` = 2^(32) + 3^(16)[2(-(1)/(2))]+[2(-(1)/(2))]`
`= 2^(32) - 3^(16) - 1`