Binomial Sum

Longish solution, can't seem to find a better one.

$(1+x)^{20} = 1 + {20 \choose 1}x + {20 \choose 2}x^2 + ... + {20 \choose 19}x^{19} + x^{20}$

$(1+\omega x)^{20} = 1 + {20 \choose 1}\omega x + {20 \choose 2}\omega ^2x^2 + ... + {20 \choose 19}(\omega^{19}) x^{19} + \omega^{20} x^{20}$

$(1+\omega^2x)^{20} = 1 + {20 \choose 1}\omega^2x + {20 \choose 2}(\omega^2)^2 x^2 + ... + {20 \choose 19}(\omega^2)^{19}x^{19} + (\omega^2)^{20}x^{20}$

Where $\omega , \omega^2$ are the complex cube roots of unity.Adding all three we get,

$(1+x)^{20} + (1+\omega x)^{20} + (1+\omega^2x)^{20} = 3{20 \choose 3}x^3 + 3{20 \choose 6}x^6 + ... + 3{20 \choose 18}x^{18}$

Since $1^n + \omega^n + (\omega^2)^n = 0$ if $3 \not | n$ and $1^n + \omega^n + (\omega^2)^n = 3$ if $3 | n$

Differentiating w.r.t. $x$

$20((1+x)^{19} + \omega (1+\omega x)^{19} + \omega^2(1+\omega^2x)^{19}) = 3( 3{20 \choose 3}x^2 + 6{20 \choose 6}x^5 + ... + 18{20 \choose 18}x^{17})$

Again Differentiating w.r.t. $x$ .

$20*19((1+x)^{18} + \omega^2 (1+\omega x)^{18} + \omega(1+\omega^2x)^{18}) = 3( 3*2{20 \choose 3}x + 6*5{20 \choose 6}x^4 + ... + 18*17{20 \choose 18}x^{16})$

Adding both the above equations and putting $x = 1$ , we have,

$20((2)^{19} + \omega (1+\omega)^{19} + \omega^2(1+\omega^2)^{19}) + 20*19((2)^{18} + \omega^2 (1+\omega )^{18} + \omega(1+\omega^2)^{18}) = 3( 3{20 \choose 3} + 6{20 \choose 6} + ... + 18{20 \choose 18} + 3*2{20 \choose 3} + 6*5{20 \choose 6} + ... + 18*17{20 \choose 18} )$

Since $1 + \omega^2 = -\omega$ and $1 + \omega = -\omega^2$

$\Rightarrow 20((2)^{19} + \omega (-\omega^2)^{19} + \omega^2(-\omega)^{19}) + 20*19((2)^{18} + \omega^2 (-\omega^2)^{18} + \omega(-\omega)^{18}) = 3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

Since $(-\omega)^{6n} = (-\omega)^{6n} = 1$ , we have

$\Rightarrow 20((2)^{19} + \omega (-\omega^2) + \omega^2(-\omega)) + 20*19((2)^{18} + \omega^2 + \omega) =3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

And $\omega + \omega^2 = -1$

$\Rightarrow 20((2)^{19} + (-1) + (-1)) + 20*19((2)^{18} +(-1)) =3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

$\Rightarrow 20((2)^{19} - 2) + 20*19((2)^{18} -1)) =3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

$\Rightarrow 40((2)^{18} -1) + 380((2)^{18} - 1) =3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

$\Rightarrow 420((2)^{18} -1) =3( 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18} )$

$\Rightarrow 140((2)^{18} -1) = 3^2{20 \choose 3} + 6^2{20 \choose 6} + ... + 18^2{20 \choose 18}$ .

Therefore $k = \boxed{140}$