Question

Let $f_1$ be a differentiable function satisfying $f_2(x) = \lim_{n \to \infty} \binom{n}{k} \left( f_1(\frac{x}{n}) - f_1(0) \right)^k; \qquad f_1'(0) = \frac{1}{3},$ $f_3(k) = \lim_{x \to 0} \frac{1}{x} \sum_{r=1}^{k} f_3(\frac{x}{r}); \qquad k \in N, f_3(0) = 0, f_3'(0) = 1.$ Then which of the following option(s) is/are equal to $\lambda = \sum_{r=0}^{10} \left( \binom{10}{r} \cdot r! \cdot f_2(3) + \frac{f_3(r+1)-f_3(r)}{f_3(11)} \right).$

• A. 1024
• B. 1026
• C. 1025
• D. 1203

Answer 1

Answer: (C)

1025

Answer 2

The problem requires us to evaluate a complex sum $\lambda$ by first determining the functions $f_2(x)$ and $f_3(k)$.

Step 1: Evaluate $f_2(x)$

We are given $f_2(x) = \lim_{n \to \infty} \binom{n}{k} \left( f_1(\frac{x}{n}) - f_1(0) \right)^k$ and $f_1'(0) = \frac{1}{3}$. Since $f_1$ is differentiable, we know that $\lim_{h \to 0} \frac{f_1(h) - f_1(0)}{h} = f_1'(0)$. Let $h = \frac{x}{n}$. As $n \to \infty$, $h \to 0$. So, $\lim_{n \to \infty} \frac{f_1(\frac{x}{n}) - f_1(0)}{\frac{x}{n}} = f_1'(0) = \frac{1}{3}$.

We can rewrite the term $\left( f_1(\frac{x}{n}) - f_1(0) \right)^k$ as: $\left( f_1(\frac{x}{n}) - f_1(0) \right)^k = \left( \frac{f_1(\frac{x}{n}) - f_1(0)}{\frac{x}{n}} \cdot \frac{x}{n} \right)^k = \left( \frac{f_1(\frac{x}{n}) - f_1(0)}{\frac{x}{n}} \right)^k \left(\frac{x}{n}\right)^k$ As $n \to \infty$, this approaches $\left(f_1'(0)\right)^k \left(\frac{x}{n}\right)^k = \left(\frac{1}{3}\right)^k \frac{x^k}{n^k}$.

Now consider the binomial coefficient $\binom{n}{k}$: $\binom{n}{k} = \frac{n(n-1)\dots(n-k+1)}{k!} = \frac{n^k \left(1-\frac{1}{n}\right)\dots\left(1-\frac{k-1}{n}\right)}{k!}$ As $n \to \infty$, $\frac{n(n-1)\dots(n-k+1)}{n^k} \to 1$. So, $\binom{n}{k} \approx \frac{n^k}{k!}$ for large $n$.

Substitute these into the expression for $f_2(x)$: $f_2(x) = \lim_{n \to \infty} \left( \frac{n^k}{k!} \right) \left( \left(\frac{1}{3}\right)^k \frac{x^k}{n^k} \right)$ $f_2(x) = \frac{x^k}{k! 3^k}$ In the summation for $\lambda$, $f_2(3)$ is used, and the summation index is $r$. This implies $k$ in $f_2(x)$ should be replaced by $r$ when used in the summation. So, $f_2(3) = \frac{3^r}{r! 3^r} = \frac{1}{r!}$.

Step 2: Evaluate $f_3(k)$

We are given $f_3(k) = \lim_{x \to 0} \frac{1}{x} \sum_{r=1}^{k} f_3(\frac{x}{r})$ with $f_3(0) = 0$ and $f_3'(0) = 1$. Let $S(x) = \sum_{r=1}^{k} f_3(\frac{x}{r})$. As $x \to 0$, $S(x) \to \sum_{r=1}^{k} f_3(0) = \sum_{r=1}^{k} 0 = 0$. The limit is of the form $\frac{0}{0}$, so we can apply L'Hopital's rule: $f_3(k) = \lim_{x \to 0} \frac{\frac{d}{dx} \left( \sum_{r=1}^{k} f_3(\frac{x}{r}) \right)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\sum_{r=1}^{k} f_3'(\frac{x}{r}) \cdot \frac{1}{r}}{1}$ As $x \to 0$, $\frac{x}{r} \to 0$, so $f_3'(\frac{x}{r}) \to f_3'(0) = 1$. Therefore, $f_3(k) = \sum_{r=1}^{k} f_3'(0) \cdot \frac{1}{r} = \sum_{r=1}^{k} 1 \cdot \frac{1}{r} = \sum_{r=1}^{k} \frac{1}{r}$ This is the $k$-th harmonic number, $H_k$. So, $f_3(k) = H_k$.

Step 3: Calculate $\lambda$

We need to calculate $\lambda = \sum_{r=0}^{10} \left( \binom{10}{r} \cdot r! \cdot f_2(3) + \frac{f_3(r+1)-f_3(r)}{f_3(11)} \right)$.

Let's evaluate the terms inside the summation: First term: $\binom{10}{r} \cdot r! \cdot f_2(3)$ Using $f_2(3) = \frac{1}{r!}$: $\binom{10}{r} \cdot r! \cdot \frac{1}{r!} = \binom{10}{r}$

Second term: $\frac{f_3(r+1)-f_3(r)}{f_3(11)}$ Using $f_3(k) = H_k$: $f_3(r+1) - f_3(r) = H_{r+1} - H_r = \left( H_r + \frac{1}{r+1} \right) - H_r = \frac{1}{r+1}$ The denominator is $f_3(11) = H_{11}$. So the second term is $\frac{\frac{1}{r+1}}{H_{11}} = \frac{1}{(r+1)H_{11}}$.

Now substitute these back into the expression for $\lambda$: $\lambda = \sum_{r=0}^{10} \left( \binom{10}{r} + \frac{1}{(r+1)H_{11}} \right)$ We can split the summation into two parts: $\lambda = \sum_{r=0}^{10} \binom{10}{r} + \sum_{r=0}^{10} \frac{1}{(r+1)H_{11}}$

Part 1: $\sum_{r=0}^{10} \binom{10}{r}$ This is the sum of binomial coefficients for $n=10$, which is $2^{10}$. $\sum_{r=0}^{10} \binom{10}{r} = 2^{10} = 1024$

Part 2: $\sum_{r=0}^{10} \frac{1}{(r+1)H_{11}}$ Since $H_{11}$ is a constant with respect to $r$, we can factor it out: $\frac{1}{H_{11}} \sum_{r=0}^{10} \frac{1}{r+1}$ Let $j = r+1$. When $r=0$, $j=1$. When $r=10$, $j=11$. The sum becomes: $\sum_{j=1}^{11} \frac{1}{j} = H_{11}$ So, the second part of the sum is: $\frac{1}{H_{11}} \cdot H_{11} = 1$

Final value of $\lambda$: $\lambda = 1024 + 1 = 1025$

The final answer is $\boxed{\text{1025}}$.