To find the value of $a$ in the expansion of $(x + 1)^7$, where there is a term of the form $ax^3$, we can use the binomial theorem. The binomial theorem gives the expansion of $(x + 1)^7$ as:

$(x + 1)^7 = \sum_{k=0}^{7} \binom{7}{k} x^k$

Here, $\binom{7}{k}$ represents the binomial coefficient, which is calculated as:

$\binom{7}{k} = \frac{7!}{k!(7-k)!}$

We need to find the term where $x^k = x^3$. This occurs when $k = 3$.

The corresponding term in the expansion is given by:

$\binom{7}{3} x^3$

Now, we calculate $\binom{7}{3}$:

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Thus, the term is $35x^3$, which means $a = 35$.

Answer:

$a = 35$.

Would you like further details or have any questions?

Here are 5 follow-up questions to explore further:

1. How does the binomial theorem apply to larger exponents, such as $(x + 2)^{10}$?
2. Can you find the general form for the $x^2$ term in $(x + y)^7$?
3. How would the result change if the power was negative, like in $(x + 1)^{-7}$?
4. What is the relationship between binomial coefficients and Pascal's triangle?
5. How does the value of $a$ change if the exponent changes, say from $(x + 1)^7$ to $(x + 1)^8$?

Tip: When expanding binomials, identifying the desired term (like $x^k$) early can help speed up calculations!