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

Question

Quizplus · Accepted Answer

We can write the given sum as:
$$3(1+2+3+\ldots+n)$$
Using the formula for sum of first n natural numbers, we get:
$$3\left(\frac{n(n+1)}{2}\right)=\frac{3n(n+1)}{2}$$
Therefore, option C is the correct choice. To prove that this is true, we can use mathematical induction. 
Base case: For n=1, the sum is 3, which is equal to $\frac{3\cdot 1(1+1)}{2}$.
Inductive step: Assume that the statement is true for some arbitrary natural number k. i.e.,
$$3 + 6 + 9 \ldots 3k=\frac{3k(k+1)}{2}$$
We need to prove that it holds for k+1. i.e., we need to show that:
$$3 + 6 + 9 \ldots 3(k+1)=\frac{3(k+1)(k+2)}{2}$$
We know that:
$$3 + 6 + 9 \ldots 3(k+1)=3 + 6 + 9 \ldots 3k + 3(k+1)=\frac{3k(k+1)}{2}+3(k+1)=\frac{(k+1)(3k+6)}{2}=\frac{3(k+1)(k+2)}{2}$$
This completes the proof.

Select the Right Expression for the Following Sum and Prove nnn

Select the Right Expression for the Following Sum and Prove $n$