If aⁿ = Σᵣ₌₀ⁿ 1/nCr, then Σᵣ₌₀ⁿ r/nCr equals (a) (n - 1)aⁿ (b) naⁿ (c) (1/2)naⁿ (d) None of the abov

Q: Why can't we just use the formula for $\sum r \cdot ^nC_r$?

That formula is for the product, not the reciprocal! The reciprocal $\frac{1}{^nC_r}$ changes the behavior entirely. We must use symmetry because the denominator is symmetric, not the numerator.

Q: Why does replacing $r$ with $n-r$ not change the sum?

The summation index $r$ is a dummy variable. Whether you sum from $r=0$ to $n$ or from $n$ down to $0$, the set of terms being added is identical. It is like reading a book from the last page to the first; the story remains the same.

Q: Is this trick only for binomial coefficients?

It is a general technique for any symmetric sum. Whenever you see a denominator that is symmetric, look for a way to use that symmetry to simplify the numerator. It is a powerful tool in your JEE arsenal.

Question

If $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$, then $\sum_{r=0}^n \frac{r}{^nC_r}$ equals

Options:

(A) $(n - 1)a_n$

(B) $na_n$

(C) $\frac{1}{2}na_n$

(D) None of the above

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the sum $S = \sum_{r=0}^n \frac{r}{^nC_r}$, given the sequence $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$. The presence of the variable $r$ in the numerator often acts as a roadblock in binomial summations.

However, in JEE mathematics, when you encounter a structure that exhibits symmetry, the most efficient approach is to leverage that symmetry rather than attempting direct algebraic expansion.

### The Obstacle and the Insight

The variable $r$ is trapped in the numerator, changing with every step of the summation. This prevents us from treating it as a constant.

We rely on the fundamental property of binomial coefficients:
$$^nC_r = ^nC_{n-r}$$

This is not merely a formula; it is a geometric reality. It implies that the distribution of binomial coefficients is perfectly symmetric, meaning the values are identical whether we count from the beginning or the end of the sequence.

### The Mathematical Dance

Because of this symmetry, we can perform a transformation by replacing the index $r$ with $(n-r)$. The total value of the sum $S$ remains invariant under this substitution:
$$S = \sum_{r=0}^n \frac{n-r}{^nC_{n-r}}$$

Applying the symmetry property $^nC_{n-r} = ^nC_r$, we rewrite the expression as:
$$S = \sum_{r=0}^n \frac{n-r}{^nC_r}$$

We can now split this fraction into two distinct parts:
$$S = \sum_{r=0}^n \frac{n}{^nC_r} - \sum_{r=0}^n \frac{r}{^nC_r}$$

### The Final Collapse

Observe the structure we have created. The first term contains a constant $n$ that can be factored out of the summation:
$$S = n \sum_{r=0}^n \frac{1}{^nC_r} - \sum_{r=0}^n \frac{r}{^nC_r}$$

Recognizing that $\sum_{r=0}^n \frac{1}{^nC_r}$ is our original sequence $a_n$ and the second term is our original sum $S$, the equation simplifies to:
$$S = n \cdot a_n - S$$

Adding $S$ to both sides yields $2S = n \cdot a_n$. Therefore, the final value of the sum is:
**$$S = \frac{n}{2} a_n$$**

We have conquered the problem by recognizing the hidden symmetry. Keep this technique in your toolkit—whenever you see a symmetric denominator, look for the mirror image to simplify the expression.

Animated Solution for Mathematics - Binomial Theorem: If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ , then $\sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ equals

Select Answer:

Visualized Solution

Given Sequence $a_{n}$

The Target Sum $S$

Identifying the Obstacle

Symmetry of Binomial Coefficients

Reversing the Index

Simplifying the Denominator

Splitting the Numerator

Distributing the Sum Operator

Factoring out $n$

Substituting $a_{n}$ and $S$

The Linear Equation in $S$

Final Calculation

Final Answer

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Obstacle and the Insight

The Mathematical Dance

The Final Collapse

Similar Questions

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

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

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series $\frac{2 ( \frac{n}{2} )! ( \frac{n}{2} )!}{n !} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - .... + (- 1)^{n} (n + 1) C_{n}^{2}]$ , where $n$ is an even positive integer, is equal to

Find the sum of the series : $\sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} [\frac{1}{2 ^{r}} + \frac{3 ^{r}}{2 ^{2 r}} + \frac{7 ^{r}}{2 ^{3 r}} + \frac{1 5 ^{r}}{2 ^{4 r}} \dots up to m terms]$ .

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ is equal to:

The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + .... +^{20} C_{10}$ is

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 :

If $\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$ , then $α$ is equal to :

Let $α = \sum_{k = 0}^{n} \frac{( ^{n} C _{k} ) ^{2}}{k + 1}$ and $β = \sum_{k = 0}^{n - 1} \frac{^{n} C _{k}^{n} C _{k + 1}}{k + 2}$ . If $5 α = 6 β$ , then $n$ equals

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 an​=∑r=0n​nCr​1​, then ∑r=0n​nCr​r​ equals

Select Answer:

Visualized Solution

Given Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

The Target Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S