Question

$f\left(n\right)=\sum _{k=1}^n\sum _{j=k}^n{(}^n{C}_j\left){(}^j{C}_k\right)$ then sum of digits of $f\left(6\right)$ is

Answer 1

We expand $f\left(n\right)$.

$f\left(n\right)=C_1^n{C}_1^1+C_2^n{C}_1^2+C_3^n{C}_1^3+......+C_n^n{C}_1^n$.......(k=1)

$+C_2^n{C}_2^2+C_3^n{C}_2^3+C_4^n{C}_2^4+......+C_n^n{C}_2^n$..........(k=2)

$+C_3^n{C}_3^3+C_4^n{C}_3^4+C_5^n{C}_3^5+......+C_n^n{C}_3^n$..........(k=3)

$+...$

$C_n^n{C}_n^n$........(k=n)

$f\left(n\right)=C_1^n\left(C_1^1\right)+C_2^n(C_1^2+C_2^2)+C_3^n(C_1^3+C_2^3+C_3^3)+......+C_n^n(C_1^n+C_2^n+....+C_n^n)$

We know that $C_1^a+C_2^a+....+C_a^a=2^a-1$

$\therefore f\left(n\right)=C_1^n(2^1-1)+C_2^n(2^2-1)+C_3^n(2^3-1)+.......+C_n^n(2^n-1)$

$=C_1^n\left(2^1\right)+C_2^n\left(2^2\right)+C_3^n\left(2^3\right)+.......+C_n^n\left(2^n\right)-(C_n^1+C_n^2+C_n^3+.......+C_n^n)$

We know $(x+y{)}^n=C_0^n(x^n)+C_1^n(x^{n-1})y+C_2^n(x^{n-2}{y}^2)+.......+C_n^n(y^n)$

Here $x=1$ and $y=2$

$\therefore f\left(n\right)=\left(\right(1+2{)}^n-1)-(2^n-1)=3^n-2^n$

$f\left(6\right)=3^6-2^6=729-64=665$

Sum of digits is $=6+6+5=17$