Question

What is the sum of the series (image below) a. 1,830 b. 2,010 c. 5,400 d. 5,490

Answer 1

Answer:

d.  5490

Step-by-step explanation:

$\begin{aligned}\displaystyle \sf \sum_{n=1}^{60} 3n & =3(1)+3(2)+3(3)+...+3(49)+3(50)\\ & = 3+6+9+...+147+150\end{aligned}$

Therefore, this is an arithmetic series with:

• first term (a) = 3
• common difference (d) = 3

Sum of the first n terms of an arithmetic series:

$S_n=\dfrac12n[2a+(n-1)d]$

Therefore, the sum of the first 60 terms:

$\begin{aligned}S_{60} &=\dfrac12(60)[2(3)+(60-1)3]\\ & =30[6+(59)3]\\ & = 30[6+177]\\ & = 30[183]\\ & = 5490 \end{aligned}$