Proving that $3+11 +19 + \cdots + (8n-5)= 4n^2 -n$ by induction

Question

$ 3+11 +19 + \cdot \cdot \cdot + (8n-5)= 4n^2 -n.$

I rewrote the problem in sigma notation. Would this change the answer?

$$ \sum_{i=1}^n (8n-5) = 4n^2-n.$$

$\color{red}{Proof :}$

(i) $\;$ The statement is true for $n=1$ because

$\left( 8(1)-5) = 4(1)^2-1 \right) \Rightarrow 3=3$

(ii) $\;$ Let $n =k $ $\Rightarrow \; \text{Induction Hypothesis} $$$\sum_{i=1}^{k} (8k-5) = 4(k)^2-k$$

$\text{We need to show :}$ $ n= k+1$

$$\sum_{i=1}^{k+1} \left( 8(k+1)-5 \right) = 4(k+1)^2-(k+1)$$

$= \left( 4(k)^2-k \right) + \left( 8(k+1)-5 \right) \Rightarrow 4k^2+7k+3$

Thus the statement is true for $n+1$

(iii) By the principle of PMI, statement is true for every $n \in ℕ$.

Yet here is my question. If I write this question in sigma notation would the answer still be correct. Or do I have to write this answer without it. Would this notation be fine?

The rewrote in sigma notation has a typo that changes the meaning. It should be $\sum_{i=1}^n (8i-5)$. Variants of that typo occur later. — André Nicolas, Jul 06 '16 at 03:33
Would this go at the very beginning of the proof? I just thought that there would be no i since there was no i when I wrote the problem. Why do we need the i? what is the i? — Jon, Jul 06 '16 at 03:46
@LittleJon You may find the post on how to write a clear induction proof to be of some use, especially if you plan on constructing more inductive proofs in the future. — Daniel W. Farlow, Jul 06 '16 at 03:55
The $i$ is the index of summation. So for example to find $\sum_{i=1}^5 i^2$ you put $i=1, 2, 3, 4, 5$, calculate $i^2$ and add up. — André Nicolas, Jul 06 '16 at 04:13

score 4 · Accepted Answer · answered Jul 06 '16 at 03:53

As Andre Nicolas points out, your primary issue occurs in how you write your summation. If you do it correctly, then your work should look, more or less, as follows (side bar comments removed...see if you can follow one line to the next): \begin{align} \sum_{i=1}^{k+1}(8i-5)&= \sum_{i=1}^k(8i-5)+8(k+1)-5\\[1em] &= (4k^2-k)+8(k+1)-5\\[1em] &= 4k^2+7k+3\\[1em] &= 4k^2+8k+4-k-1\\[1em] &= 4(k+1)^2-(k+1). \end{align}

score 2 · Answer 2 · answered Jul 06 '16 at 04:20

Let $S(n)$ be the statement: $3+11+19+\cdots+(8n-5)=4n^{2}-n$

Basis step: $S(1)$:

LHS: $\big(8(1)-5\big)=3$

RHS: $4(1)^{2}-(1)=3$

$\hspace{55 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)

Inductive step:

Assume $S(k)$ is true, i.e. assume that $3+11+19+\cdots+(8k-5)=4k^{2}-k$

For $S(k+1)$:

LHS: $\hspace{7 mm}\underline{3+11+19+\cdots+(8k-5)}+\big(8(k+1)-5\big)$

$\hspace{12 mm}=4k^{2}-k+8k+8-5$

$\hspace{12 mm}=4k^{2}+7k+3$

$\hspace{12 mm}=4k^{2}+4k+3k+3$

$\hspace{12 mm}=4k\hspace{1 mm}(k+1)+3\hspace{1 mm}(k+1)$

$\hspace{12 mm}=(k+1)(4k+3)$

RHS: $\hspace{6 mm}4\hspace{1 mm}(k+1)^{2}-(k+1)$

$\hspace{12 mm}=4\hspace{1 mm}(k^{2}+2k+1)-k-1$

$\hspace{12 mm}=4k^{2}+8k+4-k-1$

$\hspace{12 mm}=4k^{2}+7k+3$

$\hspace{12 mm}=4k^{2}+4k+3k+3$

$\hspace{12 mm}=4k\hspace{1 mm}(k+1)+3\hspace{1 mm}(k+1)$

$\hspace{12 mm}=(k+1)(4k+3)$

$\hspace{55 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)

So, $S(k+1)$ is true whenever $S(k)$ is true.

Therefore, $3+11+19+\cdots+(8n-5)=4n^{2}-n$.

Essentially what I missed to include in my proof is verifying both sides of the equation. I left that out therefore my proof was incomplete. So for all inductive proofs one must make sure that LHS = RHS. — Jon, Jul 06 '16 at 17:58
Yes, we need to show that LHS $=$ RHS. My method involved simplifying the expression on the LHS, and then arriving at the same result using the RHS, but Daniel W. Farlow's method (from LHS to RHS) is valid as well. — Tazwar, Jul 06 '16 at 18:32

Proving that $3+11 +19 + \cdots + (8n-5)= 4n^2 -n$ by induction

2 Answers2