91Ó°ÊÓ

We have to prove that

$H_1\left(n\right)=n$

$H_1\left(n\right)=n$refers to the ways in which one number $x_i$is selected from $n$positive integers $1$through $n$.

We know that, $1$number can be selected from$n$different numbers in$n$ways.

Therefore,role="math" localid="1648218917857" $H_1\left(n\right)=n$.

We have to prove that $H_k\left(n\right)=\underset{j=1}{\overset{n}{∑}}{H}_{k-1}\left(j\right)k>1$

For k=1, we have proved that $H_1\left(n\right)=n$.

if k=2.

$H_2\left(n\right)$is the no. of ways of selecting any two positive numbers from 1 through n, as we know this can happen in two ways:

That is we have two numbers $\left(x_1,x_2\right)$from $2$numbers.

Here, $x_2$is the maximum number. $x_2$is either $1\text{or}2$as there are only two members.

For $x_2=1$, we need to select one number from $(2-1)$numbers from $1$through $1$, we denote it as $H_1\left(1\right)$.

For $x_2=2$, we need to select one number from $(2-1)$numbers from $1$through $2$, we denote it as $H_1\left(2\right)$.

So, for n = 2, $H_2\left(2\right)=H_1\left(1\right)+H_1\left(2\right)$

Therefore, $H_2\left(n\right)=\underset{j=1}{\overset{n}{∑}}{H}_{2-1}\left(j\right);k>1$

So, for $H_k\left(n\right)=H_{k-1}\left(1\right)+H_{k-1}\left(1\right)+........+H_{k-1}\left(n\right)$

Hence it is proved that localid="1648220403818" $H_k\left(n\right)=\underset{j=1}{\overset{n}{∑}}{H}_{k-1}\left(j\right)k>1$

We have, $H_k\left(n\right)=\underset{j=1}{\overset{n}{∑}}{H}_{k-1}\left(j\right)k>1$

So,$H_3\left(5\right)=\underset{j=1}{\overset{5}{∑}}{H}_{3-1}\left(j\right)k>1$$H_3\left(5\right)=\underset{j=1}{\overset{5}{∑}}{H}_{2-1}\left(j\right)k>1$

localid="1648229266636" $H_3\left(5\right)=H_2\left(1\right)+H_2\left(2\right)+H_2\left(3\right)+H_2\left(4\right)+H_2\left(5\right)$

$H_2\left(1\right)=H_2\left(1\right)=1$

$H_2\left(2\right)=H_2\left(1\right)+H_2\left(1\right)=1+2=3$

localid="1648220276713" $H_2\left(3\right)=H_2\left(1\right)+H_2\left(2\right)+H_2\left(3\right)=1+2+3=6$

$$\begin{gathered} H_2\left(4\right)=H_2\left(1\right)+H_2\left(2\right)+H_2\left(3\right)+H_2\left(4\right)=1+2+3+4=10 \\ {H}_2\left(5\right)=H_2\left(1\right)+H_2\left(2\right)+H_2\left(3\right)+H_2\left(4\right)+H_2\left(5\right)=1+2+3+4+5=15 \end{gathered}$$

Therefore,$H_3\left(5\right)=1+3+6+10+15=35$