Question

Prove using mathematical induction that for all $n\ge 1$

1+4+7+..+(3n-2)=$\frac{n(3n-1)}{2}$

• A. I want to see the solution
• B. Take me to next question

Answer 1

The correct option is B Take me to next question

For any integer $n\ge 1$,$P_n$ be the statement that

1+4+7+..+(3n-2)=$\frac{n(3n-1)}{2}$

$\underset{–}{Basecase}$: The statement $P_1$ says that

$1=\frac{1(3-1)}{2}$.

Which is true.

Inductive step. Fix $k\ge 1$ , and supoose that $P_k$ holds, that is,

$1+4+7+...+(3k-2)=\frac{k(3k-1)}{2}$.

It remains to show that $p_{k+1}$ holds, that is,

$1+4+7+...+\left(3\right(k+1)-2)=\frac{(k+1)\left(3\right(k+1)-1)}{2}$.

$1+4+7+...+\left(3\right(k+1)-2)=1+4+7+...+\left(3\right(k+1)-2)$

= $1+4+7+...+(3k+1)$

= $1+4+7+...+(3k-2)+(3k+1)$

=$\frac{k(3k-1)}{2}+(3k+1)$

= $\frac{k(3k-1)+2(3k+1)}{2}$

=$\frac{3k^2-k+6k+2}{2}$

=$\frac{3k^2+5k+2}{2}$

=$\frac{(k+1)(3k+2)}{2}$

= $\frac{(k+1)\left(3\right(k+1)-1)}{2}$

$\therefore P_{k+1}holds$.

thus, by the principle of mathematical induction for all $n\ge 1$,$P_n$ holds.