If f(x) = xⁿ, then the value of f(1) - f'(1)/1! + f''(1)/2! - f'''(1)/3! + ... + (-1)ⁿ fⁿ(1)/n! is (

Q: Wait, why do we suddenly introduce factorials? Where did they come from?

The factorials arise naturally from the repeated application of the power rule. When you differentiate $x^n$ repeatedly, you generate a product of descending integers: $n(n-1)(n-2)...$. By multiplying and dividing by $(n-k)!$, we complete the factorial $n!$, which is the elegant bridge connecting derivatives to binomial coefficients.

Q: Why can't we just calculate the first few terms and guess the pattern?

While pattern recognition is a powerful tool in the exam hall, JEE Advanced rewards the rigorous 'why.' By deriving the general term $\binom{n}{k}$, we prove the result for any $n$, ensuring you don't fall into traps where the pattern might break or change for specific values.

Q: Does this result hold for all $n$?

Yes, absolutely! As long as $n$ is a non-negative integer, the Binomial Theorem holds perfectly. The beauty of this problem is that it collapses to $(1-1)^n$, which is $0^n$, and as long as $n \ge 1$, the result is a clean, satisfying zero.

Question

If $f(x) = x^n$, then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + \dots + \frac{(-1)^n f^n(1)}{n!}$ is

Options:

(A) $1$

(B) $2^n$

(C) $2^n - 1$

(D) $0$

Sigma Percentile · Accepted Answer

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of derivatives and factorials.

You see an expression like $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \dots + \frac{(-1)^n f^{(n)}(1)}{n!}$ and your instinct might be to panic. But I want you to take a deep breath.

In the world of JEE Advanced, intimidation is just a mask for a hidden, elegant simplicity. Let us peel back that mask.

### The Anatomy of the Function

We begin with the function $f(x) = x^n$. Our goal is to evaluate the series:

$$S = \sum_{k=0}^n (-1)^k \frac{f^{(k)}(1)}{k!}$$

To understand this, we must understand the behavior of the derivatives of $x^n$. Let us calculate the first few.

The first derivative is $f'(x) = nx^{n-1}$. At $x=1$, this is simply $n$.

The second derivative is $f''(x) = n(n-1)x^{n-2}$. At $x=1$, this is $n(n-1)$. Do you see the rhythm? It is a dance of descending integers.

### The Generalization

Now, let us generalize for the $k$-th derivative. Following the pattern:

$$f^{(k)}(x) = n(n-1)(n-2)\dots(n-k+1)x^{n-k}$$

When we evaluate this at $x=1$, the $x$ term vanishes into unity. We are left with the product $n(n-1)\dots(n-k+1)$.

To make this mathematically beautiful, we multiply and divide by $(n-k)!$. This transforms our product into:

$$\frac{n!}{(n-k)!}$$

This is the moment where the logic of the problem aligns with the algebra.

### The Binomial Connection

Look at the general term of our series: $\frac{f^{(k)}(1)}{k!}$. Substituting our result, we get:

$$\frac{n!}{k!(n-k)!}$$

Does that look familiar? It is the definition of the binomial coefficient $\binom{n}{k}$.

Our entire intimidating series is actually just the alternating sum of binomial coefficients:

$$S = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n}$$

This is the heart of the problem.

### The Final Collapse

We have arrived at the final act. Recall the Binomial Theorem:

$$(1-x)^n = \sum_{k=0}^n \binom{n}{k} (-x)^k$$

If we expand this, we get $\binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 - \dots + (-1)^n \binom{n}{n}x^n$.

Now, look at our series $S$ and look at this expansion. They are identical if we set $x=1$.

Therefore, our series $S$ is equal to $(1-1)^n$. And what is $(1-1)^n$? It is $0^n$, which is **0**.

The complexity vanishes, leaving behind a beautiful, clean **0**. This is the power of recognizing patterns and connecting disparate concepts. Keep practicing, keep questioning, and you will master this.

Animated Solution for Mathematics - Binomial Theorem: If $f (x) = x^{n}$ , then the value of $f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + \dots + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$ is

Select Answer:

Visualized Solution

Analyze the Function $f (x) = x^{n}$

Calculate the First Derivative $f^{'} (1)$

Calculate the Second Derivative $f^{''} (1)$

Generalize the $k$ -th Derivative $f^{(k)} (1)$

Relate to Binomial Coefficients $(k n)$

Rewrite the Original Series

Apply the Binomial Theorem

Final Calculation and Conclusion

The Sigma Insight: Properties of Binomial Coefficients

The Anatomy of the Function

The Generalization

The Binomial Connection

The Final Collapse

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xn, then the value of f(1)−1!f′(1)​+2!f′′(1)​−3!f′′′(1)​+⋯+n!(−1)nfn(1)​ is