1. ## Recurrence relation

Solve the recurrence relation
$h_n = h_{n-1} + n^3$
with the initial condition $h_0 = 0$, and verify your solution by induction on n.

2. Didn't you forget some $h_{n+1}$ ? :/

'Cause your relation is true just for $n=1$ no matter the value of $h_1$

Hugal

3. Ok ! I got what you meant. It is :

$\displaystyle h_n = h_{n-1}+n^3$

right ?

So, now just try to figure out what the first terms of this sequence are.
Try to find $h_1, h_2, h_3, h_4$ and you'll have quite a good idea of what actually the sequence is.
After that, you'll have to prove this. Induction is an easy way to make it.
Good luck,

Hugal

4. Originally Posted by nurdynik
Solve the recurrence relation
$h_n = h_{n-1} + n^3$
with the initial condition $h_0 = 0$, and verify your solution by induction on n.
See this webpage.

5. Originally Posted by nurdynik
Solve the recurrence relation
$h_n = h_{n-1} + n^3$
with the initial condition $h_0 = 0$, and verify your solution by induction on n.
The solution is well known...

$\displaystyle h_{n} = \sum_{k=0}^{n} k^{3} = \frac{n^{2}\ (n+1)^{2}}{4}$

Kind regards

$\chi$ $\sigma$

6. Hello, nurdynik!

Solve the recurrence relation: . $h_n \:=\: h_{n-1} + n^3,\;\;h_0 = 0$
and verify your solution by induction on $n.$

Crank out the first few terms and you may see a pattern.

. . $\begin{array}{cc}
n & h_n \\ \hline
0 & 0 \\ 1 & 1 \\ 2 & 9 \\ 3 & 36 & 4 & 100 \\ 5 & 225 \\ \vdots & \vdots \end{array}$

We see that the terms of the sequence are squares,
. . not every consecutive square,
. . but squares of certain numbers.

. . $\begin{array}{ccccc}
0 &=& 0^2 &=& (0)^2 \\
1 &=& 1^2 &=& (0\!+\!1)^2 \\
9 &=& 3^2 &=& (0\!+\!1\!+\!2)^2 \\
36 &=& 6^2 &=& (0\!+\!1\!+\!2\!+\!3)^2\\
100 &=& 10^2 &=& (0\!+\!1\!+\!2\!+\!3\!+\14)^2 \\
\vdots && \vdots && \vdots \end{array}$

These are the squares of Triangular Numbers, starting with $T_0 = 0.$

$\text{Therefore: }\;h_n \;=\;\left[\dfrac{n(n+1)}{2}\right]^2$