JEE Main 2026 (23 January Shift 1)

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Animated Solution for Mathematics - Binomial Theorem: The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

Select Answer:

Visualized Solution

Identify the Series and General Term

The given series is: $S = \frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$
We can write this in summation notation as:
$S = \sum_{r = 50}^{100} \frac{^{100} C _{r}}{r + 1}$

Introduce the Binomial Identity

We use the standard binomial identity:
$\frac{^{n} C _{r}}{r + 1} = \frac{1}{n + 1}^{n + 1} C_{r + 1}$

Apply Identity to the General Term

Substitute $n = 100$ into the identity:
$\frac{^{100} C _{r}}{r + 1} = \frac{1}{101}^{101} C_{r + 1}$

Rewrite the Summation

Substitute the new general term back into the sum:
$S = \sum_{r = 50}^{100} (\frac{1}{101}^{101} C_{r + 1})$
Pull out the constant factor:
$S = \frac{1}{101} \sum_{r=50}^{100} ^{101}C_{r+1}$

Change of Variable for Index

Let $k = r + 1$ .
When $r = 50$ , $k = 51$ .
When $r = 100$ , $k = 101$ .
The sum becomes: $S = \frac{1}{101} \sum_{k=51}^{101} ^{101}C_k$

Total Sum of Binomial Coefficients

Recall the property: $\sum_{k=0}^{n} ^nC_k = 2^n$
For $n = 101$ , the total sum is:
$\sum_{k=0}^{101} ^{101}C_k = 2^{101}$

Symmetry of Binomial Coefficients

Using symmetry: $^{n} C_{k} =^{n} C_{n - k}$
Since $n = 101$ is odd, the total number of terms is $102$ (even).
Sum of first half = Sum of second half
$\sum_{k=0}^{50} ^{101}C_k = \sum_{k=51}^{101} ^{101}C_k$

Calculate the Half-Sum

Total Sum = Sum of First Half + Sum of Second Half
$2^{101} = 2 \times \sum_{k=51}^{101} ^{101}C_k$
Therefore, $\sum_{k=51}^{101} ^{101}C_k = \frac{2^{101}}{2} = 2^{100}$

Final Result and Conclusion

Substitute the value of the sum back into the expression for $S$ :
$S = \frac{1}{101} \times 2^{100}$
Final Answer: $S = \frac{2 ^{100}}{101}$
This corresponds to Option (4).

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The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

When you first look at the expression

S = \frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}

it is natural to feel a slight tremor of hesitation. It looks like a daunting sum of fractions, but in the world of JEE Advanced, complexity is often a mask for elegance.

We begin by identifying the general term. We can write this series as:

S = r = 50 \sum 100 \frac{^{100} C _{r}}{r + 1}

The Secret Identity

The secret weapon for this problem is the identity:

\frac{^{n} C _{r}}{r + 1} = \frac{1}{n + 1}^{n + 1} C_{r + 1}

This is a beautiful manipulation of factorials. When you expand

^{n} C_{r}

, you get

\frac{n !}{r ! ( n - r )!}

. Dividing by

r + 1

gives you

\frac{n !}{( r + 1 )! ( n - r )!}

If you multiply and divide by

n + 1

, you suddenly see the structure of

^{n + 1} C_{r + 1}

emerge. It is like finding a hidden door in a wall you thought was solid.

Transforming the Sum

By applying this identity to our sum, we transform the expression into:

S = \frac{1}{101} \sum_{r=50}^{100} ^{101}C_{r+1}

Suddenly, the denominator is gone, replaced by a constant factor outside the sum. We perform a change of variable, letting

k = r + 1

Our sum becomes:

S = \frac{1}{101} \sum_{k=51}^{101} ^{101}C_k

Final Calculation

Now, we invoke the symmetry of binomial coefficients. We know that the total sum of the row is:

\sum_{k=0}^{101} ^{101}C_k = 2^{101}

Because

101

is odd, the total number of terms is

102

, which is an even number. The sum of the first half equals the sum of the second half.

Therefore, the sum of the second half is:

\sum_{k=51}^{101} ^{101}C_k = \frac{2^{101}}{2} = 2^{100}

Finally, we multiply by our constant

\frac{1}{101}

to arrive at the elegant result: $S = \frac{2 ^{100}}{101}$

The fear was just a lack of perspective. Keep this identity in your toolkit, and you will conquer any binomial series that dares to cross your path.

Common Traps & FAQs

Wait, why can't we just expand the combinations? There are only 51 terms, right?

While technically possible, expanding 51 terms of binomial coefficients like

^{100} C_{50}

would be a computational nightmare and a massive waste of time in a JEE exam. The beauty of this problem lies in recognizing the pattern and using the identity

\frac{^{n} C _{r}}{r + 1} = \frac{1}{n + 1}^{n + 1} C_{r + 1}

to collapse the series into a manageable form. Always look for the elegant path!

Why does the sum change from

r = 50

100

k = 51

101

That is a great observation! We performed a change of variable, setting

k = r + 1

. When

r = 50

k

becomes

50 + 1 = 51

. When

r = 100

k

becomes

100 + 1 = 101

. This shift is necessary to align our summation index with the new binomial coefficient

^{101} C_{k}

How do we know the sum of the first half equals the second half?

This comes from the symmetry property of binomial coefficients:

^{n} C_{k} =^{n} C_{n - k}

. For

n = 101

, the total number of terms is

102

(from

k = 0

101

). Because

102

is even, we have two perfectly balanced halves: the first half (

k = 0

50

) and the second half (

k = 51

101

). Since

^{n} C_{k} =^{n} C_{n - k}

, every term in the first half has a corresponding equal term in the second half.

Animated Solution for Mathematics - Binomial Theorem: The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

Select Answer:

Visualized Solution

Identify the Series and General Term

Introduce the Binomial Identity

Apply Identity to the General Term

Rewrite the Summation

Change of Variable for Index

Total Sum of Binomial Coefficients

Symmetry of Binomial Coefficients

Calculate the Half-Sum

Final Result and Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Secret Identity

Transforming the Sum

Final Calculation

Similar Questions

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The value of the expression $^{47} C_{4} + \sum_{j = 1}^{5}^{52 - j} C_{3}$ is equal to

$\sum_{k = 0}^{6}^{51 - k} C_{3}$ is equal to

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

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$ . If the sum of $(n + 1)$ terms $^{n} C_{0}, 3 \cdot^{n} C_{1}, 5 \cdot^{n} C_{2}, 7 \cdot^{n} C_{3}, \dots$ is equal to $2^{100} \cdot 101$ , then $2 [\frac{n - 1}{2}]$ is equal to ___

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ where $q$ is a real number and $q \neq = 1$ . If $^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ then $α$ is equal to:

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 51100C50​​+52100C51​​+⋯+101100C100​​ is :

Select Answer:

Visualized Solution

Identify the Series and General Term

Introduce the Binomial Identity

Apply Identity to the General Term

Rewrite the Summation

Change of Variable for Index

Total Sum of Binomial Coefficients

Symmetry of Binomial Coefficients

Calculate the Half-Sum

Final Result and Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Secret Identity

Transforming the Sum

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Cr​=25Cr​ and C0​+5⋅C1​+9⋅C2​+⋯+101⋅C25​=225⋅k then k is equal to__________.

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Animated Solution for Mathematics - Binomial Theorem: The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

The value of the expression $^{47} C_{4} + \sum_{j = 1}^{5}^{52 - j} C_{3}$ is equal to

$\sum_{k = 0}^{6}^{51 - k} C_{3}$ is equal to

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

The value $\sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ is