Induction proof of $1 + 6 + 11 +\cdots + (5n-4)=n(5n-3)/2$ I need help getting started with this proof.
Prove using mathematical induction.
$$
1 + 6 + 11 + \cdots + (5n-4)=n(5n-3)/2 
$$
$$
n=1,2,3,...
$$
I know for my basis step I need to set $n=1$ but I can't get past that. Thank you for any help.
 A: $$p(1): 1=\frac{1(5(1)-3)}{2}\\p(k):1+6+11+...+(5n-4)=\frac{n(5n-3)}{2}\\$$now try to prove $$p(k+1):1+6+11+...+(5n-4)+(5(n+1)-4)=\frac{(n+1)(5(n+1)-3)}{2}\\$$
A: First, show that this is true for $n=1$:


*

*$\sum\limits_{i=1}^{1}5i-4=\dfrac{5-3}{2}$


Second, assume that this is true for $n$:


*

*$\sum\limits_{i=1}^{n}5i-4=\dfrac{n(5n-3)}{2}$


Third, prove that this is true for $n+1$:


*

*$\sum\limits_{i=1}^{n+1}5i-4=\left(\sum\limits_{i=1}^{n}5i-4\right)+5(n+1)-4$

*$\left(\sum\limits_{i=1}^{n}5i-4\right)+5(n+1)-4=\dfrac{n(5n-3)}{2}+5(n+1)-4$ assumption used here

*$\dfrac{n(5n-3)}{2}+5(n+1)-4=\dfrac{(n+1)(5n+2)}{2}$

*$\dfrac{(n+1)(5n+2)}{2}=\dfrac{(n+1)(5n+5-3)}{2}$

*$\dfrac{(n+1)(5n+5-3)}{2}=\dfrac{(n+1)(5(n+1)-3)}{2}$
A: So you need to prove $$ \sum_{k=1}^n (5k-4) = n(5n-3)/2 $$ by induction.
You start by checking that the formula works for some $n$ ($n=1$ for example)
$$ n=1 \Rightarrow \sum_{k=1}^1 (5k-4) = (5-4) = 1 = (5-3)/2 $$
which is true.
Then your goal is to prove that the formula holding for $n$ implies that it holds for $n+1$. Start by writing the $n+1$ case using the $n$ case
$$ \sum_{k=1}^{n+1} (5k-4) = \sum_{k=1}^{n} (5k-4) + 5(n+1) - 4 $$
and substitute the formula for the $n$ case (known to hold for some $n$)
$$ \sum_{k=1}^{n+1} (5k-4) = n(5n-3)/2 + 5(n+1) - 4 .$$
The proof is complete if the last form can be shown to be equal to $(n+1)(5(n+1)-3)/2$. (Because you have shown that it holds for $n=1$ and that if it holds for some $n$ it holds also for $n+1$, therefore it holds for all $n \geq 1$.)
A: You have the induction assumption $$1+2+\ldots+(5(n-1)-4)=\frac{(n-1)(5(n-1)-3)}{2}$$
Add $5n-4$ to that... And you should get $P(n-1)\Rightarrow P(n)$ and this completes the induction
A: $$1=1(5\times 1-3)/2$$
$$1+6+11+\cdots+(5k-4)=^?k(5k-3)/2$$
$$1+6+11+\cdots+(5k-4)+(5k+1)=^?(k+1)(5k+2)/2$$
$$k(5k-3)/2+(5k+1)=^?(k+1)(5k+2)/2$$
$$k(5k-3)/2+(5k+1)=^?(k+1)(5k+2)/2$$
After some algebra...
$$(5k^2+7k+2)/2==(5k^2+7k+2)/2$$
which is true.
A: A Proof Without Using Induction:
$$\begin{align}LHS&=1+6+11+\cdots +(5n-4)\\&=[1+(5\times 1-5)]+[1+(5\times 2-5)]+\cdots +[1+(5\times n-5)]\\&=n+5\cdot\frac{n(n+1)}{2}-5n\\&=n+5\cdot\frac{n(n-1)}{2}\\&=\frac{n}{2}\cdot(2+5n-5)\\&=\frac{n(5n-3)}{2}\\&=RHS\end{align}$$
A: I will outline a simplified version that is more drawn out but probably answers your question more clearly. Your goal is to prove that the statement $P(n)$, that is,
$$
P(n) : 1+6+11+\cdots+(5n-4)=\frac{n(5n-3)}{2}
$$
holds for all $n\geq 1$. 
Base step: As you noted, check the $n=1$ case for the base step. Using $n=1$, we have that
$$
1=\frac{1(5(1)-3)}{2},
$$
which is correct. Thus, the base step checks out. 
Inductive step: Fix $k\geq 1$ and assume that
$$
P(k) : 1+6+11+\cdots+(5k-4)=\frac{k(5k-3)}{2}
$$
holds. It remains to show that
$$
P(k+1) : 1+6+11+\cdots+(5k-4)+[5(k+1)-4]=\frac{(k+1)[5(k+1)-3]}{2}
$$
follows. Starting with the left-hand side of $P(k+1)$,
\begin{align}
1+\cdots+(5k-4)+[5(k+1)-4] &\leq \frac{k(5k-3)}{2}+[5(k+1)-4]\tag{ind. hyp.}\\[1em]
                           &=\frac{k(5k-3)+10(k+1)-8}{2}\tag{com. dom.}\\[1em]
                           &=\frac{5k^2-3k+10k+10-8}{2}\tag{expand}\\[1em] 
                           &=\frac{5k^2+7k+2}{2}\tag{simplify}\\[1em]
                           &=\frac{(k+1)(5k+2)}{2}\tag{factor}\\[1em]
                           &=\frac{(k+1)[5(k+1)-3]}{2}\tag{manipulate}
\end{align}
one arrives at the right-hand side of $P(k+1)$, thus completing the inductive step.
Thus, by mathematical induction, $P(n)$ is true for all $n\geq 1$. 
