The image contains two mathematical exercises:

1. Finding the term independent of $x$ in an expansion: This part involves determining the term in a binomial expansion that does not contain $x$ (the constant term). Based on the calculations shown, the answer seems to be $184,756$.

2. Exercise 3.2: This exercise asks for the coefficient of $x^7$ in the expansion of $\left( x^2 + \frac{7}{x} \right)^8$.

Solution to Exercise 3.2

For the binomial expansion $\left( x^2 + \frac{7}{x} \right)^8$, we are looking for the term where the power of $x$ equals 7.

In the general term for a binomial expansion $(a + b)^n$, the $r$-th term is given by: $T_{r+1} = \binom{n}{r} \cdot a^{n-r} \cdot b^r$

In this case:

• $a = x^2$
• $b = \frac{7}{x}$
• $n = 8$

The general term becomes: $T_{r+1} = \binom{8}{r} \cdot (x^2)^{8-r} \cdot \left(\frac{7}{x}\right)^r$

Simplifying, we have: $T_{r+1} = \binom{8}{r} \cdot x^{2(8-r)} \cdot \frac{7^r}{x^r} = \binom{8}{r} \cdot 7^r \cdot x^{16 - 3r}$

For this term to contain $x^7$, we set the power of $x$ equal to 7: $16 - 3r = 7$ $3r = 9$ $r = 3$

So, the required term is when $r = 3$: $T_{4} = \binom{8}{3} \cdot 7^3 \cdot x^7$

Calculating this: $\binom{8}{3} = \frac{8!}{3!(8-3)!} = 56$ $7^3 = 343$ $T_4 = 56 \cdot 343 \cdot x^7 = 19208 \cdot x^7$

Thus, the coefficient of $x^7$ is 19,208.

Would you like further details on any of these steps?

Related Questions

1. How can we determine the term independent of $x$ in any binomial expansion?
2. What is the general approach for finding the coefficient of a specific power in binomial expansions?
3. Can you explain how binomial coefficients are calculated and their significance?
4. How does the general formula for a term in a binomial expansion apply to other types of expressions?
5. What other methods exist for finding terms in an expansion without expanding the entire expression?

Tip

For problems involving powers of $x$, setting up an equation for the exponent (like $16 - 3r = 7$) helps pinpoint the exact term quickly, making these types of problems faster to solve.