Prove by induction that $1+4+7+...+(3n-2) = 2n(3n-1)$ I have an exercise where I, using induction, have to prove the following:
\begin{equation*}
1 + 4 + 7 + \ldots + (3n-2) = 2n(3n-1).
\end{equation*}
I immediately got stuck on the base case with $n=1$ because the following should be true: $1 = 2 \cdot 1 \cdot (3 \cdot 1 - 1) = 2 \cdot 2$ which is clearly not the case. However the exercise says to prove the given relation, not to check if it is correct, which is why I have elected to make this post to see if I am missing any obvious thing.   
 A: First of all, the claim is wrong. A direct proof shows that.
The sequence $\{1,4,7,\ldots,3n-2\}$ is an arithmetic progression with common difference $3$ and there are $n$ number of terms in it. So, the sum of all the terms is:
$$\begin{align}S =1+4+7+\ldots+(3n-2)&=1+(1+3)+(1+6)+\ldots+(1+3(n-1))&\\ &=\sum_{i=1}^n(1)+\sum_{i=1}^{n-1}(3i)\\&=n+3\cdot\frac{(n-1)n}{2}=\frac{3n^2-n}{2}=\frac{n(3n-1)}{2}\neq 2n(3n-1)\end{align}$$
The "$\neq$" follows in general. Now, if you were to prove the corrected claim using induction, here's a hint for the inductive step:
$$\begin{align}S_{n+1} & =1+4+7+\ldots+(3n-2)+(3n+1)\\
&=S_n+(3n+1)\\
&=\frac{n(3n-1)}{2}+(3n+1)\tag{by I.H.}\\&=\frac{3n^2-n+6n+2}{2}\\&=\frac{3n^2+5n+2}{2}\\&=\frac{(n+1)(3n+2)}{2}\\
& =\ldots\end{align}$$
A: You can always repeat Gauss:
$
\phantom{2}S = 1 + 4 + 7 + \cdots + (3n-2)
$
$
\phantom{2}S = (3n-2) + (3n-5) + (3n-8) + \cdots + 1
$
$
2S = (3n-1) + (3n-1) + (3n-1) + \cdots + (3n-1) = n(3n-1).
$
A: $\displaystyle f_1(n) = \sum_{k\,=\,0}^{n-1} 3k\!+\!1,\ $ $\,f_2(n) = \dfrac{n(3n\!-\!1)}2\,$ are both solutions of $\,f(n\!+\!1)-f(n) = 3n\!+\!1,\,$ with equal initial condition $\,f(1)=1,\,$ so induction implies they remain equal for all $\,n,\,$ since the induction step is clear: $\, f_1(n\!+\!1) = f_1(n) + 3n\!+\!1 = f_2(n) + 3n\!+\!1 = f_2(n\!+\!1)$
Remark $\ $ This uniqueness theorem is a special case of telescopy. Further examples can be found in prior posts.
