Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =( 4(4n + 1)(8n + 7)) / 6

I've only gotten this far: [4⋅6 + 5 * 7 + 6 * 8 + ... + 4n(4n + 2)] + 4(n + 1)(8n + 7)

2. 2. 1^2 + 4^2 + 7^2 + ... + (3n – 2)^2 = (n(6n^2 - 3n - 1)) / 2

3. For the given statement Pn, write the statements P1, Pk, and Pk+1.
2 + 4 + 6 + . . . + 2n = n(n+1)
I have no idea how to properly do these problems. Please show me the whole process of solving these problems, and the answers please! Thanks!

Answer 1

Establish the base case. For \(n=1\),
\[\begin{align*}4\cdot6&=\frac{4(4+1)(8+7)}{6}\\
24&\not=50\end{align*}\]
Always check the base case. The first statement is false.

Answer 2

For the second statement, check the base case (again: ALWAYS check the base case, it's the foundation of induction proofs). For \(n=1\),
\[\begin{align*}1^2&=\frac{1(6-3-1)}{2}\\
1&=1
\end{align*}\]
Now assume the statement holds for \(n=k\), and use this assumption to show that it holds for \(n=k+1\).

In other words, you assume the following is true:
\[1^2+4^2+\cdots+(3k-2)^2=\frac{k(6k^2-3k-1)}{2}\]
and use this to show
\[1^2+4^2+\cdots+(3k-2)^2+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
From the assumption, you can simplify \(k+1\) statement.
\[\frac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2=\frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]
From here it's a matter of algebraic rearrangement.

For future reference, consider putting forth some effort on your part. Asking for a detailed answer in full for all of your homework is unreasonable. At the very least you could have taken the time to look up a tutorial or worked example. There are plenty of those available online.

Answer 3

Understand the fundamental steps to induction:
\[\textbf{Establish the base case}\\
\quad\text{Show truth for }n=1\text{ (or whatever the starting index might be).}\]
\[\textbf{Prove induction hypothesis}\\
\quad\text{Assume truth for }n=k.\\
\quad\text{Show truth for }n=k+1.\]