JEE Main 2023 (01 February Shift 1)

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Animated Solution for Mathematics - Binomial Theorem: 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

Q: I forgot the property for the sum of odd binomial coefficients. Is there a quick way to derive it?

Absolutely! Start with the expansion of $(1+x)^n = \sum {}^{n}C_r x^r$. If you set $x=1$, you get the sum of all coefficients ($2^n$). If you set $x=-1$, you get the alternating sum ($0$). By subtracting the second equation from the first, the even terms cancel out, and you are left with twice the sum of the odd terms. Dividing by $2$ gives you $2^{n-1}$. Never memorize blindly; understand the derivation!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Look closely at the denominators. For the first few terms, the sum of the numbers inside the factorials is constant. Calculate

1 + 50

3 + 48

, and

5 + 46

. Do you see a pattern? Now, look at the very last term. Does it fit the pattern? If not, remember that

1!

is the same as

0!

Hint 2

Hint 2: You are dealing with a sum that screams for the Binomial Theorem. Recall the definition of a combination:

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

. To transform your series into a sum of combinations, you need a common numerator. What value should you multiply and divide the entire expression by to make every term look like a combination?

Hint 3

Hint 3: Once you have transformed the series into the form

\frac{1}{51 !} \times (^{51} C_{1} +^{51} C_{3} + \dots +^{51} C_{51})

, you are almost there. Recall the property of the sum of binomial coefficients with odd indices:

^{n} C_{1} +^{n} C_{3} +^{n} C_{5} + \dots = 2^{n - 1}

. Apply this to your specific case where

n = 51

Visualized Solution

The Given Series

Let the given sum be $S$ .
$S = \frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$

Observing the Denominators

Look at the sum of the numbers inside the factorials for each term.
Term 1: $1 + 50 = 51$
Term 2: $3 + 48 = 51$
Term 3: $5 + 46 = 51$

The Anomaly in the Last Term

Let's check the last term: $\frac{1}{51 ! 1 !}$
Sum of numbers: $51 + 1 = 52$
This breaks the pattern of $51$ !

Fixing the Pattern

Recall that $1! = 1$ and $0! = 1$ .
Therefore, $51! 1! = 51! \times 1 = 51! 0!$
Rewrite the last term: $\frac{1}{51 ! 0 !}$
Now, $51 + 0 = 51$ . The pattern is restored!

The Multiplication Trick

We need a common numerator to form binomial coefficients.
Since the sum of denominators is $51$ , we need $51!$ in the numerator.
Multiply and divide the entire series $S$ by $51!$ .

Distributing $51!$

$S = \frac{1}{51 !} \times [51! (\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \dots + \frac{1}{51 ! 0 !})]$
Distribute $51!$ to every term inside the bracket:
$S = \frac{1}{51 !} [\frac{51 !}{1 ! 50 !} + \frac{51 !}{3 ! 48 !} + \dots + \frac{51 !}{51 ! 0 !}]$

Recall Binomial Coefficient Formula

The formula for combinations is:
$^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$
Notice that $r + (n - r) = n$ .
This perfectly matches our terms where the denominator sum equals the numerator!

Converting to $^{n} C_{r}$

Apply the formula to our terms:
$\frac{51 !}{1 ! 50 !} =^{51} C_{1}$
$\frac{51 !}{3 ! 48 !} =^{51} C_{3}$
$\frac{51 !}{51 ! 0 !} =^{51} C_{51}$

The Simplified Series

Substitute these back into the equation for $S$ :
$S = \frac{1}{51 !} [^{51} C_{1} +^{51} C_{3} +^{51} C_{5} + \dots +^{51} C_{51}]$
The bracket now contains the sum of binomial coefficients with odd lower indices.

Sum of Odd Binomial Coefficients

Recall the standard binomial property:
$^{n} C_{1} +^{n} C_{3} +^{n} C_{5} + \dots = 2^{n - 1}$
This is a favorite concept of JEE!

Evaluating the Bracket

In our series, $n = 51$ .
Sum inside the bracket $=^{51} C_{1} +^{51} C_{3} + \dots +^{51} C_{51}$
Apply the formula: $2^{51 - 1} = 2^{50}$

Final Answer

Substitute the bracket value back into $S$ :
$S = \frac{1}{51 !} \times 2^{50}$
$S = \frac{2 ^{50}}{51 !}$
This matches option (1).

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The expression provided is:

S = \frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}

At first glance, the denominators appear to follow a pattern where the sum of the factorial arguments is

51

. However, the final term

\frac{1}{51 ! 1 !}

yields a sum of

52

To restore symmetry, we utilize the identity

0! = 1

. By rewriting the final term as

\frac{1}{51 ! 0 !}

, the sum of the arguments becomes

51 + 0 = 51

, perfectly aligning with the rest of the series.

The Binomial Bridge

To relate this to the Binomial Theorem, we recall the definition of a combination:

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}

Since our constant sum is

51

, we multiply and divide the entire series by

51!

to introduce the necessary numerator:

S = \frac{1}{51 !} k \in {1, 3, \dots, 51} \sum \frac{51 !}{k ! ( 51 - k )!}

This transformation allows us to express the series in terms of binomial coefficients:

S = \frac{1}{51 !} (^{51} C_{1} +^{51} C_{3} +^{51} C_{5} + \dots +^{51} C_{51})

The Elegant Conclusion

We are now evaluating the sum of binomial coefficients with odd lower indices for

n = 51

. A fundamental property of binomial coefficients states that the sum of odd-indexed terms is given by

2^{n - 1}

Substituting

n = 51

into this property:

k odd \sum^{51} C_{k} = 2^{51 - 1} = 2^{50}

Multiplying this by the factor

\frac{1}{51 !}

kept outside the summation, we arrive at the final result:

$S = \frac{2 ^{50}}{51 !}$

Common Traps & FAQs

Wait, why did we change

1!

0!

in the last term? Isn't that cheating?

It is not cheating; it is mathematical insight! In mathematics, we often look for symmetry. The pattern of the denominators summing to

51

was broken by the last term. Since

0! = 1

and

1! = 1

, replacing

1!

with

0!

is perfectly valid. It restores the symmetry required to apply the Binomial Theorem. Always look for ways to make the structure consistent!

Why did we multiply by

51!

specifically? How would I know to do that in an exam?

That is a great question. In JEE Advanced, when you see a sum of fractions involving factorials, your goal is to create the structure of a binomial coefficient:

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

. By observing that the sum of the indices in the denominator was

51

, we knew that

n

must be

51

. Therefore, we needed

51!

in the numerator to complete the definition of

^{51} C_{r}

. It is about recognizing the 'missing piece' of the puzzle.

I forgot the property for the sum of odd binomial coefficients. Is there a quick way to derive it?

Absolutely! Start with the expansion of

(1 + x)^{n} = \sum^{n} C_{r} x^{r}

. If you set

x = 1

, you get the sum of all coefficients (

2^{n}

). If you set

x = - 1

, you get the alternating sum (

0

). By subtracting the second equation from the first, the even terms cancel out, and you are left with twice the sum of the odd terms. Dividing by

2

gives you

2^{n - 1}

. Never memorize blindly; understand the derivation!

Animated Solution for Mathematics - Binomial Theorem: 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

Select Answer:

Visualized Solution

The Given Series

Observing the Denominators

The Anomaly in the Last Term

Fixing the Pattern

The Multiplication Trick

Distributing $51!$

Recall Binomial Coefficient Formula

Converting to $^{n} C_{r}$

The Simplified Series

Sum of Odd Binomial Coefficients

Evaluating the Bracket

Final Answer

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Binomial Bridge

The Elegant Conclusion

Similar Questions

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

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

Let $S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots$ up to 13 terms. If $13 S = \frac{2 ^{k}}{n !}, k \in N$ , then $n + k$ is equal to

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

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

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

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

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

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

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 1!50!1​+3!48!1​+5!46!1​+⋯+49!2!1​+51!1!1​ is

Select Answer:

Visualized Solution

The Given Series

Observing the Denominators

The Anomaly in the Last Term

Fixing the Pattern

The Multiplication Trick

Distributing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 51!

Recall Binomial Coefficient Formula

Converting to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nCr​

The Simplified Series

Sum of Odd Binomial Coefficients

Evaluating the Bracket

Final Answer

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Binomial Bridge

The Elegant Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3 is

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Animated Solution for Mathematics - Binomial Theorem: 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

Distributing $51!$

Converting to $^{n} C_{r}$

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

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

Let $S = \frac{1}{25 !} + \frac{1}{3 ! 23 !} + \frac{1}{5 ! 21 !} + \dots$ up to 13 terms. If $13 S = \frac{2 ^{k}}{n !}, k \in N$ , then $n + k$ is equal to

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

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

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

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

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

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