Let’s work through the problems in the image step by step. Since there are multiple problems, I’ll solve them systematically. I'll start with the Binomial Theorem section. Let’s begin:

Problem 1:

Find the value of r if (2r + 3)ᵗʰ and (r − 1)ᵗʰ terms in the expansion of (1 + x)¹⁵ are equal.

Solution:

The general term in the expansion of $(1+x)^{15}$ is: $T_{r+1} = \binom{15}{r} x^r$

For the terms $(2r+3)^{\text{th}}$ and $(r-1)^{\text{th}}$, their coefficients must be equal: $\binom{15}{2r+2} = \binom{15}{r-2}$

Using the property of binomial coefficients: $\binom{n}{k} = \binom{n}{n-k}$

This simplifies to: $2r+2 + r-2 = 15$

Solve for $r$: $3r = 15 \implies r = 5$

Answer: $r = 5$

Problem 2:

Find the coefficient of $x^{-7}$ in the expansion of $\left(ax - \frac{1}{bx^2}\right)^{11}$.

Solution:

The general term in the expansion of $\left(ax - \frac{1}{bx^2}\right)^{11}$ is: $T_{r+1} = \binom{11}{r} (ax)^{11-r} \left(-\frac{1}{bx^2}\right)^r$

Simplify: $T_{r+1} = \binom{11}{r} a^{11-r} (-1)^r b^{-r} x^{11-r-2r}$

The power of $x$ is: $11 - r - 2r = 11 - 3r$

We need the coefficient of $x^{-7}$: $11 - 3r = -7 \implies 3r = 18 \implies r = 6$

Now substitute $r = 6$ into the coefficient expression: $\text{Coefficient} = \binom{11}{6} a^{11-6} (-1)^6 b^{-6}$

$= \binom{11}{6} a^5 b^{-6}$

$\binom{11}{6} = 462$

Thus, the coefficient is: $462a^5/b^6$

Answer: $462a^5/b^6$

Problem 3:

Find the term which is independent of $x$ in the expansion of $\left(x^2 + \frac{1}{x}\right)^{12}$.

Solution:

The general term in the expansion of $\left(x^2 + \frac{1}{x}\right)^{12}$ is: $T_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(\frac{1}{x}\right)^r$

Simplify: $T_{r+1} = \binom{12}{r} x^{2(12-r) - r}$

The power of $x$ is: $24 - 2r - r = 24 - 3r$

The term is independent of $x$ when: $24 - 3r = 0 \implies r = 8$

Substitute $r = 8$ into the coefficient expression: $T_{9} = \binom{12}{8} (x^2)^4 \left(\frac{1}{x}\right)^8$

$T_{9} = \binom{12}{8}$

$\binom{12}{8} = 495$

Answer: $495$

Problem 4:

For what value of $r$, the coefficient of $x^r$ and $x^{r+1}$ are equal in the expansion of $(1+x)^{11}$.

Solution:

The general term is: $T_{r+1} = \binom{11}{r} x^r$

The coefficient of $x^r$ is: $\binom{11}{r}$

The coefficient of $x^{r+1}$ is: $\binom{11}{r+1}$

Equating the coefficients: $\binom{11}{r} = \binom{11}{r+1}$

Using the property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$

$r = 5$

Answer: $r = 5$

Would you like me to proceed with the remaining problems, or clarify any of these steps? Let me know if you’d like further details! 😊

Follow-Up Questions:

1. Can you explain the steps for simplifying a binomial coefficient equation?
2. Why is the term independent of $x$ found by setting the exponent to zero?
3. How do properties of binomial coefficients simplify problem-solving?
4. What are some practical applications of the binomial theorem in real life?
5. Can we solve Problem 2 differently without using $r$-based substitution?

Tip:

When solving binomial problems, always simplify the general term expression before substituting values for faster computation.