1. ## Recurrence relation

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

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

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

Hugal

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

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

right ?

So, now just try to figure out what the first terms of this sequence are.
Try to find $\displaystyle h_1, h_2, h_3, h_4$ 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
$\displaystyle h_n = h_{n-1} + n^3$
with the initial condition $\displaystyle h_0 = 0$, and verify your solution by induction on n.
See this webpage.

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Hello, nurdynik!

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

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

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

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

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

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

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