# 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.

## 7 Answers

$$p(1): 1=\frac{1(5(1)-3)}{2}\\p(k):1+6+11+...+(5n-4)=\frac{n(5n-3)}{2}\\$$now try to prove $$p(k+1):1+6+11+...+(5n-4)+(5(n+1)-4)=\frac{(n+1)(5(n+1)-3)}{2}\\$$

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 $n \geq 1$.)

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

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

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

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

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

$$1=1(5\times 1-3)/2$$ $$1+6+11+\cdots+(5k-4)=^?k(5k-3)/2$$ $$1+6+11+\cdots+(5k-4)+(5k+1)=^?(k+1)(5k+2)/2$$ $$k(5k-3)/2+(5k+1)=^?(k+1)(5k+2)/2$$ $$k(5k-3)/2+(5k+1)=^?(k+1)(5k+2)/2$$

After some algebra... $$(5k^2+7k+2)/2==(5k^2+7k+2)/2$$ which is true.

A 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}

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 $n\geq 1$.