Answer
See explanation
Work Step by Step
To prove by induction that the closed‐form formula
\[
y_k \;=\; \frac{k(k+1)(2k+1)}{6}
\]
satisfies the given recursion
\[
y_1 = 1,
\quad
y_k = y_{k-1} + k^2 \quad (k \ge 2),
\]
we carry out the following steps:
---
## Base Case \( (k = 1) \)
We check that the formula holds for \(k = 1\):
\[
y_1
\;=\; \frac{1 \cdot (1+1) \cdot (2 \cdot 1 + 1)}{6}
\;=\; \frac{1 \cdot 2 \cdot 3}{6}
\;=\; 1.
\]
This matches the given initial condition \(y_1 = 1\), so the base case is verified.
---
## Inductive Step
Assume that for some integer \(n \ge 1\), the formula holds:
\[
y_n \;=\; \frac{n(n+1)(2n+1)}{6}.
\]
We want to show that it then holds for \(n+1\), i.e.,
\[
y_{n+1} \;=\; \frac{(n+1)(n+2)\bigl(2(n+1) + 1\bigr)}{6}.
\]
Using the recursive definition:
\[
y_{n+1}
\;=\; y_n + (n+1)^2.
\]
By the inductive hypothesis,
\[
y_n = \frac{n(n+1)(2n+1)}{6}.
\]
Hence,
\[
y_{n+1}
= \frac{n(n+1)(2n+1)}{6} \;+\; (n+1)^2.
\]
It is convenient to combine these terms under a common denominator:
\[
y_{n+1}
= \frac{n(n+1)(2n+1)}{6}
\;+\; \frac{6 \bigl((n+1)^2\bigr)}{6}
= \frac{n(n+1)(2n+1) + 6(n+1)^2}{6}.
\]
Factor out \((n+1)\) inside the numerator:
\[
n(n+1)(2n+1) + 6(n+1)^2
= (n+1)\bigl[n(2n+1) + 6(n+1)\bigr].
\]
Simplify inside the brackets:
\[
n(2n+1) + 6(n+1)
= 2n^2 + n + 6n + 6
= 2n^2 + 7n + 6.
\]
Factor \(2n^2 + 7n + 6\):
\[
2n^2 + 7n + 6 = (2n+3)(n+2).
\]
Therefore,
\[
y_{n+1}
= \frac{(n+1)\,(2n+3)\,(n+2)}{6}
= \frac{(n+1)(n+2)\bigl(2(n+1)+1\bigr)}{6}.
\]
This completes the inductive step, showing that if the formula holds for \(n\), it also holds for \(n+1\).
---
## Conclusion
By the principle of mathematical induction, for all integers \(k \ge 1\),
\[
\boxed{y_k = \frac{k(k+1)(2k+1)}{6}.}
\]
Thus, the closed‐form formula satisfies both the initial condition and the recursive definition.
