Question

evaluate
\\(\displaystyle\sum_{n=1}^{4} (6n + 3)\\)
\\(\circ\\) a. \\(s_n = 144\\)
\\(\circ\\) b. \\(s_n = 36\\)
\\(\circ\\) c. \\(s_n = 72\\)
\\(\circ\\) d. \\(s_n = 126\\)

Explanation:

Step1: Expand the summation

Substitute $n=1,2,3,4$ into $(6n+3)$:
When $n=1$: $6(1)+3=9$
When $n=2$: $6(2)+3=15$
When $n=3$: $6(3)+3=21$
When $n=4$: $6(4)+3=27$

Step2: Sum the terms

Add the calculated values together:
$9 + 15 + 21 + 27$

Step3: Calculate the total

Simplify the sum:
$9+15=24$, $21+27=48$, $24+48=72$

Answer:

c. $S_n=72$