Prove by induction that $\sum_{r=1}^{n} r^{3} = \frac{1}{4} n^{2} (n+1)^{2}.$ The r th term of a series is given by $\ln\left(2a^{r}\right)$ where $a$ is a positive constant. Show that the sum of the first $n$ terms of the series is given by $\dfrac{n}{4} \ln\left(16a^{n(n+1)^{2}}\right).$

1. Define the proposition and verify the base case for $n=1$.

Let $P_n$ be the statement that $\sum_{r=1}^{n} r^{3} = \frac{1}{4} n^{2} (n+1)^{2}$.

When $n=1$,
$\text{LHS} = \sum_{r=1}^{1} r^{3} = 1^{3} = 1$
$\text{RHS} = \frac{1}{4} (1)^{2} (1+1)^{2} = \frac{1}{4}(1)(4) = 1$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the inductive hypothesis.

Assume that $P_k$ is true for some positive integer $k$.
That is, $\sum_{r=1}^{k} r^{3} = \frac{1}{4} k^{2} (k+1)^{2}$.

3. Prove the inductive step for $n=k+1$.

Consider $P_{k+1}$:
$\begin{aligned} \sum_{r=1}^{k+1} r^{3} &= \sum_{r=1}^{k} r^{3} + (k+1)^{3} \\ &= \frac{1}{4} k^{2} (k+1)^{2} + (k+1)^{3} \\ &= \frac{1}{4} (k+1)^{2} \left[k^{2} + 4(k+1)\right] \\ &= \frac{1}{4} (k+1)^{2} (k^{2} + 4k + 4) \\ &= \frac{1}{4} (k+1)^{2} (k+2)^{2} \\ &= \frac{1}{4} (k+1)^{2} ((k+1)+1)^{2} \end{aligned}$
Thus, $P_{k+1}$ is true if $P_k$ is true.

4. Write the concluding statement.

Since $P_1$ is true, and $P_k$ being true implies that $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all positive integers $n$.

1. Express the general term using log laws

The $r$th term is
$u_r=\ln\!\left(2a^{r^3}\right)=\ln 2+\ln a^{r^3}=\ln 2+r^3\ln a.$

2. Sum the series term by term using part (i)

$\sum_{r=1}^{n}u_r=\sum_{r=1}^{n}\left(\ln 2+r^3\ln a\right)=n\ln 2+\left(\ln a\right)\sum_{r=1}^{n}r^3.$
Using the result of part (i), $\displaystyle\sum_{r=1}^{n}r^3=\tfrac14 n^2(n+1)^2$, so
$\sum_{r=1}^{n}u_r=n\ln 2+\frac{1}{4}n^2(n+1)^2\ln a.$

3. Recombine into the required logarithmic form

Factor out $\tfrac{n}{4}$:
$n\ln 2=\frac{n}{4}(4\ln 2)=\frac{n}{4}\ln 16,\qquad \frac14 n^2(n+1)^2\ln a=\frac{n}{4}\cdot n(n+1)^2\ln a=\frac{n}{4}\ln a^{\,n(n+1)^2}.$
Hence
$\sum_{r=1}^{n}u_r=\frac{n}{4}\left[\ln 16+\ln a^{\,n(n+1)^2}\right]=\frac{n}{4}\ln\!\left(16\,a^{\,n(n+1)^2}\right).$