Statement-1: Σᵣ₌₀ⁿ (r + 1) nCr = (n + 2) 2ⁿ⁻¹. Statement-2: Σᵣ₌₀ⁿ (r + 1) nCr xᵣ = (1 + x)ⁿ + nx(1 +

Q: Why do we differentiate? Can't we just expand the sum?

Expanding the sum is possible for small $n$, but for a general $n$, it is impossible. Differentiation is the elegant, universal tool that handles the index $r$ by bringing it down from the exponent. It is the difference between brute force and mathematical grace.

Q: Why does the summation index change from $r=0$ to $r=1$ after differentiation?

Great observation! When $r=0$, the term is $^nC_0 x^0 = 1$. The derivative of a constant is zero. So, the $r=0$ term vanishes. This is why the sum effectively starts from $r=1$.

Q: How do we know to substitute $x=1$?

Look at the target expression in Statement-1: $(n+2)2^{n-1}$. Now look at the general form in Statement-2. If you set $x=1$, the $x^r$ terms become $1^r = 1$, and the expression simplifies beautifully. It is the bridge between the general polynomial identity and the specific numerical series.

Question

Statement-1: $\sum_{r=0}^n (r + 1) ^nC_r = (n + 2) 2^{n-1}$. Statement-2: $\sum_{r=0}^n (r + 1) ^nC_r x^r = (1 + x)^n + nx(1 + x)^{n-1}$.

Options:

(A) Statement-1 is false, Statement-2 is true

(B) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

(C) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

(D) Statement-1 is true, Statement-2 is false

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating two statements involving binomial coefficients. Statement-1 asks us to verify the identity:

$$\sum_{r=0}^n (r + 1) ^nC_r = (n + 2) 2^{n-1}$$

Statement-2 provides a general polynomial identity:

$$\sum_{r=0}^n (r + 1) ^nC_r x^r = (1 + x)^n + nx(1 + x)^{n-1}$$

Our goal is to verify both and demonstrate that Statement-2 serves as the logical foundation for Statement-1.

### The Foundation: The Standard Expansion

Every great journey in binomial series begins with the standard expansion:

$$(1 + x)^n = \sum_{r=0}^n ^nC_r x^r$$

This is our bedrock. When we observe the term $(r+1)$ multiplying the binomial coefficient in Statement-2, it serves as a clear hint that the original expansion has been transformed via calculus.

### The Calculus Key: Bringing Down the Index

Whenever an index $r$ multiplies a binomial coefficient, our intuition should immediately suggest **differentiation**. Differentiating $x^r$ with respect to $x$ brings the power $r$ down as a coefficient.

Starting with our base identity $(1 + x)^n = \sum_{r=0}^n ^nC_r x^r$, we differentiate both sides with respect to $x$:

$$\frac{d}{dx} (1 + x)^n = \frac{d}{dx} \sum_{r=0}^n ^nC_r x^r$$

$$n(1 + x)^{n-1} = \sum_{r=1}^n r ^nC_r x^{r-1}$$

Note that the summation index now starts at $r=1$ because the $r=0$ term is a constant, which vanishes upon differentiation. To align the powers of $x$, we multiply both sides by $x$:

$$nx(1 + x)^{n-1} = \sum_{r=1}^n r ^nC_r x^r$$

### The Final Synthesis: Combining the Pieces

We now examine the left-hand side of Statement-2: $\sum_{r=0}^n (r + 1) ^nC_r x^r$. We can decompose this into two separate summations:

$$\sum_{r=0}^n r ^nC_r x^r + \sum_{r=0}^n ^nC_r x^r$$

The second part is our original expansion $(1 + x)^n$. The first part is the expression we derived via differentiation: $nx(1 + x)^{n-1}$. Adding these yields:

$$nx(1 + x)^{n-1} + (1 + x)^n$$

This confirms that **Statement-2 is true**.

### The Bridge: From Polynomial to Numerical

Statement-1 is simply a specific case of Statement-2. By setting $x=1$ in our verified identity, the left side becomes $\sum_{r=0}^n (r + 1) ^nC_r$, which is the sum in Statement-1.

Substituting $x=1$ into the right side of Statement-2 gives:

$$(1 + 1)^n + n(1)(1 + 1)^{n-1} = 2^n + n 2^{n-1}$$

Factoring out $2^{n-1}$, we obtain:

$$2^{n-1}(2 + n) = (n + 2) 2^{n-1}$$

This matches the expression in Statement-1 perfectly. We have successfully proven that **both statements are true** and that Statement-2 is the logical foundation for Statement-1.

Animated Solution for Mathematics - Binomial Theorem: Statement-1: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1}$ . Statement-2: $\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1}$ .

Select Answer:

Visualized Solution

Analyzing the Statements

The Fundamental Binomial Expansion

Splitting the Summation in Statement-2

Evaluating the Second Part

Using Calculus for the First Part

Differentiating the Expansion

Adjusting the Power of $x$

Verifying Statement-2

Deriving Statement-1 from Statement-2

Final Calculation for Statement-1

Conclusion and Takeaways

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Foundation

The Calculus Key

The Final Synthesis

The Bridge

Similar Questions

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If $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to

If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

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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; } Statement-1: ∑r=0n​(r+1)nCr​=(n+2)2n−1. Statement-2: ∑r=0n​(r+1)nCr​xr=(1+x)n+nx(1+x)n−1.