For question no. 5 from the image, it is as follows:

"Verify that $3 + 7 + 11 + \dots + (4n - 1) = n(2n + 1)$ by using mathematical induction."

Solution using Mathematical Induction:

Step 1: Base Case $n = 1$

When $n = 1$, the left-hand side (LHS) is: $3 = 3$ And the right-hand side (RHS) is: $n(2n + 1) = 1(2(1) + 1) = 3$ Since LHS = RHS for $n = 1$, the base case holds true.

Step 2: Induction Hypothesis

Assume that the statement is true for $n = k$, i.e., $3 + 7 + 11 + \dots + (4k - 1) = k(2k + 1)$

Step 3: Induction Step

We need to prove that the statement is also true for $n = k + 1$, i.e., $3 + 7 + 11 + \dots + (4k + 3) = (k + 1)(2(k + 1) + 1)$ Starting with the LHS for $n = k + 1$: $LHS = \left(3 + 7 + 11 + \dots + (4k - 1)\right) + (4(k + 1) - 1)$ Using the induction hypothesis: $LHS = k(2k + 1) + (4k + 3)$ Simplifying this expression: $LHS = k(2k + 1) + 4k + 3 = 2k^2 + k + 4k + 3 = 2k^2 + 5k + 3$

Now, compute the RHS for $n = k + 1$: $RHS = (k + 1)(2(k + 1) + 1) = (k + 1)(2k + 2 + 1) = (k + 1)(2k + 3)$ Expanding this: $RHS = (k + 1)(2k + 3) = 2k^2 + 3k + 2k + 3 = 2k^2 + 5k + 3$

Since LHS = RHS, the statement holds true for $n = k + 1$.

Conclusion

By the principle of mathematical induction, the given statement is true for all $n \geq 1$.

Would you like more details or have any questions on this solution?

Here are 5 related questions:

1. What is the principle of mathematical induction, and how is it applied in proofs?
2. Can mathematical induction be used for inequalities? If so, how?
3. How would you prove a formula for the sum of the first $n$ odd numbers?
4. How does this proof compare with verifying geometric series using induction?
5. Can you apply induction to prove divisibility statements (e.g., $n^2 - n$ is always even)?

Tip: In mathematical induction, always ensure the base case is correct before proceeding to the induction step.