Suppose det [ cases ∑ _ k = 0 ^ n k & ∑ _ k = 0 ^ n C _ k n k ^ 2 ∑...

As given \left|\begin{cases}\mathop{∑}_{k = 0}^{n} k & \mathop{∑}_{k = 0}^{n} C n {}_{k} {k}^{2} \\ \mathop{∑}_{k = 0}^{n} C n {}_{k} k & \mathop{∑}_{k = 0}^{n} C n {}_{k} {3}^{k}\end{cases}\right| = 0
\left(a\right)\mathop{∑}_{k = 0}^{n}k =\dfrac{n\left(n + 1\right)}{2}$<br />\left(b\right)\mathop{∑}_{k = 0}^{n}{C}_{k} n \left({k}^{2}\right) = \mathop{∑}_{k = 0}^{n}{\left({k}^{2} - k + k\right)}^{n}{C}_{k}$
$= \mathop{∑}_{k = 0}^{n}\left({k}^{2} - k\right) {C}_{k} n + \mathop{∑}_{k = 0}^{n}k {C}_{k} n$
$= \mathop{∑}_{k = 0}^{n}k\left(k - 1\right)\dfrac{n}{k} . \dfrac{n - 1}{k - 1} {C}_{k - 2} n - 2 + \mathop{∑}_{k = 0}^{n}k\dfrac{n}{k} {C}_{k - 1} n - 1$
$= n\left(n - 1\right)\mathop{∑}_{k = 0}^{n}{C}_{k - 2} n - 2 + n\mathop{∑}_{k = 0}^{n}{C}_{k - 1} n - 1$
$= n\left(n - 1\right){2}^{n - 2} + n{2}^{n - 1} = n{2}^{n - 2}\left(n - 1 + 2\right) = n\left(n + 1\right){2}^{n - 2}$
\left($$\begin{array}{ll}∴ r {C}_{r} n = n {C}_{r - 1} n - 1 \\ ∴ \mathop{∑}_{r = 0}^{n}{C}_{r} n = {2}^{n}\end{array}$$\right)
\left(c\right)\mathop{∑}_{k = 0}^{n}{C}_{k} n k = \mathop{∑}_{k = 0}^{n}k {C}_{k} n = \mathop{∑}_{k = 0}^{n}k\dfrac{n}{k} {C}_{k - 1} n - 1$<br />$= n\mathop{∑}_{k = 0}^{n}{C}_{k - 1} n - 1 = n{2}^{n - 1}$<br />\left(d\right)$\mathop{∑}_{k = 0}^{n}{C}_{k} n {3}^{k} = {C}_{0} n + {C}_{1}3 + {C}_{2} n {3}^{2} + \ldots + {C}_{n} n {3}^{n} n$<br />$= {\left(1 + 3\right)}^{n}
$= {4}^{n}$
now \left|$\begin{cases}\dfrac{n\left(n + 1\right)}{2} & n \left(n + 1\right){2}^{n - 2} \\ n{2}^{n - 1} & {4}^{n}\end{cases}$\right| = 0$</p><p>$n\left(n + 1\right){2}^{2n - 1} - {n}^{2}\left(n + 1\right){2}^{2n - 3} = 0$</p><p>${2}^{2n - 1} - n{2}^{2n - 3} = 0$<br />$n = 4$<br />Now$\mathop{∑}_{k = 0}^{4}\dfrac{{C}_{k} 4}{k + 1} = \dfrac{1}{5}\mathop{∑}_{k = 0}^{4}{C}_{k + 1} 5$<br />$\begin{cases}= \dfrac{1}{5}\left({2}^{5} - 1\right) & \left(∴ \dfrac{{C}_{r} n}{r + 1} = \dfrac{{C}_{r + 1} n + 1}{n + 1}\right)\end{cases}$<br />$\begin{cases}= \dfrac{31}{5} =6.20 & \left(∴ \mathop{∑}_{r = 1}^{n}{C}_{r} = {2}^{r} - 1 n\right)\end{cases}