Answer
**(a)** \(\displaystyle 1 + 2 + \dots + (k-1) \;=\; \frac{k(k-1)}{2}.\)
**(b)** \(\displaystyle 3 + 2 + 4 + 6 + \dots + 2n \;=\; n(n+1) + 3.\)
**(c)** \(\displaystyle 3\cdot 1 + 3\cdot 2 + \dots + 3\cdot n \;+\; n \;=\; \frac{3n(n+1) + 2n}{2} \;=\; \frac{3n^2 + 5n}{2}.\)
Work Step by Step
## (a) Sum \(1 + 2 + 3 + \dots + (k - 1)\)
If \(k\) is an integer with \(k \ge 2\), then summing from \(1\) up to \((k - 1)\) is just the usual sum formula with \(n = k - 1\). Hence
\[
1 + 2 + 3 + \dots + (k - 1)
\;=\;
\frac{(k-1)\bigl((k-1)+1\bigr)}{2}
\;=\;
\frac{(k-1)\,k}{2}.
\]
---
## (b) Sum \(3 + 2 + 4 + 6 + 8 + \dots + 2n\)
Assume \(n \ge 1\). First note that the even integers from \(2\) up to \(2n\) sum to
\[
2 + 4 + 6 + \dots + 2n
\;=\;
2\,(1 + 2 + 3 + \dots + n)
\;=\;
2 \times \frac{n(n+1)}{2}
\;=\;
n(n+1).
\]
Therefore,
\[
3 + 2 + 4 + 6 + \dots + 2n
\;=\;
3 \;+\;\bigl(2 + 4 + 6 + \dots + 2n\bigr)
\;=\;
3 + n(n+1)
\;=\;
n(n+1) + 3.
\]
---
## (c) Sum \(3\cdot 1 + 3\cdot 2 + 3\cdot 3 + \dots + 3\cdot n \;+\; n\)
A common variation (one that uses the same sum formula) is to look at
\[
3\cdot 1 + 3\cdot 2 + 3\cdot 3 + \dots + 3\cdot n \;+\; n.
\]
Factor out the 3 from the first \(n\) terms:
\[
3\cdot 1 + 3\cdot 2 + \dots + 3\cdot n
\;=\;
3\,(1 + 2 + \dots + n)
\;=\;
3 \times \frac{n(n+1)}{2}
\;=\;
\frac{3n(n+1)}{2}.
\]
Then adding \(n\) gives
\[
3\cdot 1 + 3\cdot 2 + \dots + 3\cdot n \;+\; n
\;=\;
\frac{3n(n+1)}{2} + n
\;=\;
\frac{3n(n+1) + 2n}{2}
\;=\;
\frac{n\bigl(3(n+1) + 2\bigr)}{2}
\;=\;
\frac{3n^2 + 5n}{2}.
\]
(If your part (c) is phrased slightly differently, you can adapt the same idea: rewrite any sum of multiples of \(n\) or 3 in terms of \(1+2+\dots+n\) as needed.)
