Select the right expression for the following sum and prove that your statement is true for all natural numbers n. $P _ { n } : 3 + 6 + 9 \ldots 3 n$
A) $\quad \frac { n ( 3 n + 1 ) } { 2 }$
B) $\frac { 3 n ( n + 1 ) } { 4 }$
C) $\quad \frac { n ( 3 n + 1 ) } { 4 }$
D) $\frac { 3 n ( n + 1 ) } { 2 }$
E) none of the above

Question

Quizplus · Accepted Answer

The given series is an arithmetic series where the first term $a = 3$ and the common difference $d = 3$. The sum of the first $n$ terms of an arithmetic series is given by $S_n = \frac{n}{2}[2a + (n-1)d]$. Substituting $a = 3$ and $d = 3$ into this formula gives $S_n = \frac{n}{2}[2(3) + (n-1)(3)] = \frac{n}{2}[6 + 3n - 3] = \frac{n}{2}[3n + 3] = \frac{3n(n + 1)}{2}$, which matches option D.

Select the Right Expression for the Following Sum and Prove Pn:3+6+9…3nP _ { n } : 3 + 6 + 9 \ldots 3 nPn​:3+6+9…3n

Select the Right Expression for the Following Sum and Prove $P _ { n } : 3 + 6 + 9 \ldots 3 n$