Question
Let $S U M S Q(k)$ denote the sum of the squares of the natural numbers from 1 through $\mathrm{k}$, inclusively. Write a recursive definition of $S U M S Q(k)$ and prove that $S U M S Q(k)=(k(k+1)(2 k+1)) / 6$.
Step 1
- The sum of the squares of the natural numbers from 1 through \( k \) can be defined recursively as follows: \[ SUMSQ(k) = \begin{cases} 0, & \text{if } k = 0 \\ SUMSQ(k-1) + k^2, & \text{if } k > 0 \end{cases} \] - This definition states that Show more…
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