It is given that,

$$\left| {\matrix{ {\sum\limits_{k = 0}^n k } & {\sum\limits_{k = 0}^n {{}^n{C_k}{k^2}} } \cr {\sum\limits_{k = 0}^n {{}^n{C_k}.k} } & {\sum\limits_{k = 0}^n {{}^n{C_k}{3^k}} } \cr } } \right| = 0$$

$$ \Rightarrow \left| {\matrix{ {{{n(n + 1)} \over 2}} & {n(n + 1){2^{n - 2}}} \cr {n{{.2}^{n - 1}}} & {{4^n}} \cr } } \right| = 0$$

$$ \because $$ $$\left[ \matrix{ \sum\limits_{k = 0}^n k = {{n(n + 1)} \over 2},\,\sum\limits_{k = 0}^n {{}^n{C_k}k = n{{.2}^{n - 1}}} \hfill \cr \sum\limits_{k = 0}^n {} {}^n{C_k}{k^2} = n(n + 1){2^{n - 2}}\,\,and\,\,\sum\limits_{k = 0}^n {} {}^n{C_k}{3^k} = {4^n} \hfill \cr} \right]$$

$$ \Rightarrow {{n(n + 1)} \over 2}{4^n} - {n^2}(n + 1)\,{2^{2n - 3}} = 0$$

$$ \Rightarrow {{{4^n}} \over 2} - n{{{4^{n - 1}}} \over 2} = 0$$

$$ \Rightarrow n = 4$$

$$ \therefore $$ $$\sum\limits_{k = 0}^n {} {{{}^n{C_k}} \over {k + 1}} = \sum\limits_{k = 0}^4 {} {{{}^4{C_k}} \over {k + 1}}$$

= $${1 \over 5}\sum\limits_{k = 0}^4 {} {}^5{C_{k + 1}} = {1 \over 5}({2^5} - 1)$$

$$ = {1 \over 5}(32 - 1) = {{31} \over 5} = 6.20$$