`(1 + x)^(n) = C_(0) + C_(1)x + C_(2) x^(2) + ...+ C_(n) x^(n)` ,
show that ` sum_(r=0)^(n) C_(r)^(3)` is equal to the coefficient of ` x^(n) y^(n)` in the
expansion of ` {(1+ x)(1 + y) (x + y)}^(n)` .

Explanation

`(1 + x)^(n) (y+1)^(n) (x + y)^(n) = sum_(r=0)^(n) C_(r) x^(r)`
` sum_(s=0)^(n) C_(s)y^(n-1) sum_(t=0)^(n) C_(t) x^(n-t) y^(t)` ….(i)
Since , ` C_(2)^(3)` is the coefficient of ` x^(0) y^(n-0) x^(n-0) y^(0) `
i.e., ` x^(n) y^(n) (r = s = t = 0)`
Now , ` C_(1)^(3) ` is the coefficient of ` x^(1) y^(n-1) x^(n-1) y`
i.e., ` x^(n) y^(n) (r + s = t = 1)`
And ` C_(k) ^(3) ` is the coefficient of ` x^(k) y^(n-k) x^(n-k) y^(k)`
i.e., ` x^(n) y^(n) (r = s = t = k) `
Hence , the coefficient of ` x^(n) y^(n) " in" (1 + x)^(n) (y + 1)^(n) (x + y)^(n)`
` = C_(0) ^(3) + C_(1)^(3) + C_(2) ^(3) + ...+C_(n)^(3) = sum_(r=0)^(n) C_(r)^(3)` .