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.
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
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 $\displaystyle T_0 = 0.$
$\displaystyle \text{Therefore: }\;h_n \;=\;\left[\dfrac{n(n+1)}{2}\right]^2$