Summation Notation — Question 5

Question 5

Prove that for any positive integer $n$, $$\sum_{k=1}^{n} \bigl(3k+2\bigr) = \frac{3n(n+1)}{2}+2n.$$

Question 5 - Solution

We prove the identity using properties of summations.

Start with the given sum: $$\sum_{k=1}^{n} (3k+2).$$

Split the sum using linearity: $$\sum_{k=1}^{n} (3k+2) = \sum_{k=1}^{n} 3k + \sum_{k=1}^{n} 2.$$

Factor out constants: $$\sum_{k=1}^{n} 3k = 3\sum_{k=1}^{n} k, \qquad \sum_{k=1}^{n} 2 = 2\sum_{k=1}^{n} 1.$$

Use known summation formulas: $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}, \qquad \sum_{k=1}^{n} 1=n.$$

Substitute these results: $$\sum_{k=1}^{n} (3k+2) = 3\cdot\frac{n(n+1)}{2} + 2n.$$

Thus, $$\sum_{k=1}^{n} (3k+2) = \frac{3n(n+1)}{2}+2n.$$

$$\boxed{\sum_{k=1}^{n} (3k+2)=\frac{3n(n+1)}{2}+2n}$$