We are asked to prove by induction the following identity:

$\sum_{j=0}^{n} j 3^j = \frac{3}{4} \left[ n 3^{n+1} - (n+1) 3^n + 1 \right]$

Step 1: Base Case

We begin by testing the formula for $n = 0$.

The left-hand side is: $\sum_{j=0}^{0} j 3^j = 0$ (since the sum contains only a single term, $j = 0$).

Now, substitute $n = 0$ into the right-hand side of the formula: [ \frac{3}{4} \left[ 0 \cdot 3^{0+1} - (0+1) \cdot 3^0 + 1 \right] = \frac{3}{4} \left[ 0 - 1 + 1 \right] = \frac{3}{4} \times 0 = 0 ] Since both sides are equal, the base case holds.

Step 2: Inductive Hypothesis

Now, assume the formula holds for some arbitrary $n = k$, i.e.,

$\sum_{j=0}^{k} j 3^j = \frac{3}{4} \left[ k 3^{k+1} - (k+1) 3^k + 1 \right]$

Step 3: Inductive Step

We need to prove that the formula holds for $n = k+1$, i.e., we want to show:

$\sum_{j=0}^{k+1} j 3^j = \frac{3}{4} \left[ (k+1) 3^{(k+1)+1} - ((k+1)+1) 3^{k+1} + 1 \right]$

First, note that: $\sum_{j=0}^{k+1} j 3^j = \sum_{j=0}^{k} j 3^j + (k+1) 3^{k+1}$ By the inductive hypothesis, we substitute for the first sum: $= \frac{3}{4} \left[ k 3^{k+1} - (k+1) 3^k + 1 \right] + (k+1) 3^{k+1}$

Next, simplify the right-hand side: $= \frac{3}{4} \left[ k 3^{k+1} - (k+1) 3^k + 1 \right] + \frac{4}{4} (k+1) 3^{k+1}$ $= \frac{3}{4} \left[ k 3^{k+1} - (k+1) 3^k + 1 + 4(k+1) 3^{k+1} \right]$ $= \frac{3}{4} \left[ (k+1) 3^{k+1} + 1 - (k+1) 3^k \right]$

Now, factor the terms: $= \frac{3}{4} \left[ (k+1) 3^{k+1} - (k+1) 3^k + 1 \right]$ This matches the right-hand side of the original formula for $n = k+1$: $\frac{3}{4} \left[ (k+1) 3^{k+2} - (k+2) 3^{k+1} + 1 \right]$

Since the inductive hypothesis leads us to the desired expression, the induction step is complete.

Step 4: Conclusion

By the principle of mathematical induction, the formula is true for all non-negative integers $n$.

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Would you like more details or have any questions about this proof?

Here are some additional related questions to explore:

1. How does the formula for the sum $\sum_{j=0}^{n} j 3^j$ compare to sums for other powers of $j$?
2. Can we extend this proof to sums involving other constants in place of 3?
3. How would you derive the formula for sums of the form $\sum_{j=0}^{n} j x^j$ for general $x$?
4. What happens when we try to prove a similar formula for sums involving $j^2$ or higher powers of $j$?
5. How could this inductive method be generalized to other types of summations?

Tip: Induction is a powerful tool for proving statements involving sums, sequences, and formulas for recurrence relations. It’s important to carefully check both the base case and the inductive step to ensure the validity of your argument.