1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $n \geq 1$.)

$\endgroup$
1
$\begingroup$

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

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

A Proof Without Using Induction:

$$\begin{align}LHS&=1+6+11+\cdots +(5n-4)\\&=[1+(5\times 1-5)]+[1+(5\times 2-5)]+\cdots +[1+(5\times n-5)]\\&=n+5\cdot\frac{n(n+1)}{2}-5n\\&=n+5\cdot\frac{n(n-1)}{2}\\&=\frac{n}{2}\cdot(2+5n-5)\\&=\frac{n(5n-3)}{2}\\&=RHS\end{align}$$

$\endgroup$
0
$\begingroup$

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 $n\geq 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.