To solve the expression $(2k + 1)^5 + 7$, we'll expand the binomial $(2k + 1)^5$ using the Binomial Theorem, and then simplify it.

Step 1: Apply the Binomial Theorem

The Binomial Theorem states:

$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$

For $(2k + 1)^5$, we have $a = 2k$, $b = 1$, and $n = 5$.

$(2k + 1)^5 = \sum_{r=0}^{5} \binom{5}{r} (2k)^{5-r} 1^r$

Step 2: Expand the Terms

Now, let's calculate each term in the expansion:

1. When $r = 0$:
   $\binom{5}{0} (2k)^5 = 1 \cdot (32k^5) = 32k^5$

2. When $r = 1$:
   $\binom{5}{1} (2k)^4 = 5 \cdot (16k^4) = 80k^4$

3. When $r = 2$:
   $\binom{5}{2} (2k)^3 = 10 \cdot (8k^3) = 80k^3$

4. When $r = 3$:
   $\binom{5}{3} (2k)^2 = 10 \cdot (4k^2) = 40k^2$

5. When $r = 4$:
   $\binom{5}{4} (2k)^1 = 5 \cdot (2k) = 10k$

6. When $r = 5$:
   $\binom{5}{5} (2k)^0 = 1 \cdot (1) = 1$

Step 3: Combine the Terms

Now we can combine all the terms:

$(2k + 1)^5 = 32k^5 + 80k^4 + 80k^3 + 40k^2 + 10k + 1$

Step 4: Add 7 to the Expression

Finally, adding 7:

$(2k + 1)^5 + 7 = 32k^5 + 80k^4 + 80k^3 + 40k^2 + 10k + 1 + 7$

$= 32k^5 + 80k^4 + 80k^3 + 40k^2 + 10k + 8$

Final Answer:

$(2k + 1)^5 + 7 = 32k^5 + 80k^4 + 80k^3 + 40k^2 + 10k + 8$

Would you like more details or have any questions on this? Here are some related questions:

1. How can you apply the Binomial Theorem to a similar problem?
2. What are the properties of binomial coefficients?
3. How do you factor expressions like the one above?
4. What happens if you use a different exponent (e.g., $(2k + 1)^3$)?
5. How do you solve for $k$ if the expression is set equal to a specific number?

Tip: The Binomial Theorem is a powerful tool for expanding expressions like these, and it's often used to simplify higher powers quickly!