Question

If $f\left(n\right)=\sum _{k=0}^n{(n-2k)}^2{C}_k^2$, then find $f\left(6\right)$.

Answer 1

We have
${(n-2k)}^2={(n-k)}^2+k^2-2k(n-k)$
Thus
$\sum _{k=0}^n{(n-2k)}^2{C}_k^2=\sum _{k=0}^n{(n-k)}^2{C}_{n-k}^2+\sum _{k=0}^n{k}^2{C}_k^2-2\sum _{k=0}^{n}k(n-k)C_k^2$

$\because C_k=C_{n-k}$
But, $k(n-k)C_k^2=(k.C_k)\left(\right(n-k\left)C_{n-k}\right)$
$n\left({{}^{n-1}C}_{k-1}\right)\left(n{{}^{n-1}C}_{n-k-1}\right)=n^2\left({{}^{n-1}C}_{k-1}\right)\left({{}^{n-1}C}_k\right)$

Thus,
${\sum }_{k=0}^{n}k(n-k)C_k^2=n^2{\sum }_{k=1}^{n-1}\left({{}^{n-1}C}_{k-1}\right)\left({{}^{n-1}C}_k\right)=n^2\left({{}^{2n-2}C}_n\right)$
Hence, ${\sum }_{k=0}^n{(n-2k)}^2{C}_k^2=\left(2n^2\right)\left[\frac{(2n-2)!}{{(n-1)!}^2}\frac{(2n-2)!}{n!(n-2)!}\right]=\left(2n\right)\left({{}^{2n-2}C}_{n-1}\right)$

$\therefore f\left(6\right)=12{\times }^{10}{C}_5=3024$