Prove by induction that, for all positive integers $$ n $$, $$ \sum_{r=1}^{n}\left(4 r^{3}+3 r^{2}+r\right)=n(n+1)^{3} $$

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To prove the statement by induction, we will follow these steps:

1. **Base Case**: Verify the statement for $$ n = 1 $$.
2. **Inductive Step**: Assume the statement is true for some positive integer $$ k $$, and then prove it for $$ k + 1 $$.

### Step 1: Base Case

For $$ n = 1 $$:
$$
\sum_{r=1}^{1} \left(4r^3 + 3r^2 + r\right) = 4(1)^3 + 3(1)^2 + (1) = 4 + 3 + 1 = 8
$$
Now, we check the right-hand side:
$$
1(1 + 1)^3 = 1 \cdot 2^3 = 1 \cdot 8 = 8
$$
Since both sides are equal, the base case holds.

### Step 2: Inductive Step

Assume the statement is true for $$ n = k $$:
$$
\sum_{r=1}^{k} \left(4r^3 + 3r^2 + r\right) = k(k + 1)^3
$$
We need to show that it holds for $$ n = k + 1 $$:
$$
\sum_{r=1}^{k+1} \left(4r^3 + 3r^2 + r\right) = \sum_{r=1}^{k} \left(4r^3 + 3r^2 + r\right) + \left(4(k+1)^3 + 3(k+1)^2 + (k+1)\right)
$$
Using the inductive hypothesis:
$$
= k(k + 1)^3 + \left(4(k+1)^3 + 3(k+1)^2 + (k+1)\right)
$$
Now, we simplify the right-hand side:
$$
= k(k + 1)^3 + 4(k + 1)^3 + 3(k + 1)^2 + (k + 1)
$$
Factoring out $$ (k + 1)^3 $$:
$$
= (k + 1)^3(k + 4) + 3(k + 1)^2 + (k + 1)
$$
Now, we can combine the terms:
$$
= (k + 1)^3(k + 4) + 3(k + 1)^2 + (k + 1)
$$
Factoring $$ (k + 1) $$:
$$
= (k + 1) \left( (k + 1)^2(k + 4) + 3(k + 1) + 1 \right)
$$
Now, we need to simplify $$ (k + 1)^2(k + 4) + 3(k + 1) + 1 $$:
$$
= (k + 1)^2(k + 4) + 3(k + 1) + 1
$$
Expanding $$ (k + 1)^2(k + 4) $$:
$$
= (k^2 + 2k + 1)(k + 4) = k^3 + 4k^2 + 2k^2 + 8k + k + 4 = k^3 + 6k^2 + 9k + 4
$$
Now adding $$ 3(k + 1) + 1 $$:
$$
= k^3 + 6k^2 + 9k + 4 + 3k + 3 + 1 = k^3 + 6k^2 + 12k + 8
$$
This can be factored as:
$$
= (k + 2)(k + 1)^2
$$
Thus, we have:
$$
= (k + 1)^3(k + 1 + 1)^2 = (k + 1)^3(k + 2)^3
$$
This shows that:
$$
\sum_{r=1}^{k+1} \left(4r^3 + 3r^2 + r\right) = (k + 1)(k + 2)^3
$$
Thus, the inductive step holds.

### Conclusion

By the principle of mathematical induction, the statement is true for all positive integers $$ n $$:
$$
\sum_{r=1}^{n}\left(4 r^{3}+3 r^{2}+r\right)=n(n+1)^{3}
$$
By induction, for all positive integers $$ n $$, $$ \sum_{r=1}^{n}\left(4 r^{3}+3 r^{2}+r\right)=n(n+1)^{3} $$

Prove by induction that, for all positive integers \( n \), \[ \sum_{r=1}^{n}\left(4 r^{3}+3 r^{2}+r\right)=n(n+1)^{3} \]

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