Principle of induction proof ...

Question

Using the principle of induction prove that for every $n>0$:

$$\sum_{k=1}^n (3k-2) = \frac{3n^2-n}{2}.$$

For $n=1$: $\quad\sum_{k=1}^1 (3-2) = \frac{3-1}{2}=\sum_{k=1}^1 1=1$

Now how can I prove it for $n=n+1$?

Answer 1

Without giving away the whole answer .... Start with:

$$\begin{align}\sum_{k=1}^{n+1} (3k-2) &= \sum_{k=1}^{n} (3k-2) + (3(n+1)-2) \\&\overset{\text{Inductive Hypothesis}}= \frac{3n^2-n}{2} + (3(n+1)-2) \\&= \cdots\end{align}$$

... and hopefully this works out to $$\frac{3(n+1)^2-(n+1)}{2}$$

Good luck!

Answer 2

Inductive step

Assume true $\sum_{k=1}^n (3k-2) = \frac{3n^2-n}{2}$ then

$$\begin{align}\sum_{k=1}^{n+1} (3k-2) &= \sum_{k=1}^{n} (3k-2) + (3(n+1)-2) \\&\overset{I.H.}= \frac{3n^2-n}{2} + (3n+1) \\&= \frac{3n^2-n+6n+2}{2}\\&=\frac{3(n^2+2n+1)-n-1}{2}\\&=\frac{3(n+1)^2-(n+1)}{2}\end{align}$$

Answer 3

By proving

$\sum_{k=1}^{n+1} (3k - 2) =$

$\sum_{k=1}^{n} (3k - 2) + (3(n+1) -2)=$

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

That last line should just be organ grinding. (or sausage making.... or both.)

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

$\frac{3n^2-n}{2} + 3n+1 = $

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

$\frac {3n^2 + 6n + 3 - n-1}{2}=$

$\frac {3(n+1)^2 - (n+1)}2$