The value of 100 C5051 + 100 C5152 + + 100 C100101 is: — Solution | JEE Mathematics

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

Answer

Correct answer:D

Standard Method

Given:

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

Find: The value of $S$ .

Use the identity

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

So,

S = \frac{1}{101}\left[^{101}C_{51} + ^{101}C_{52} + \dots + ^{101}C_{101}\right]

Now, the sum of all binomial coefficients for $n = 101$ is

\sum_{r=0}^{101} {}^{101}C_r = 2^{101}

Also, by symmetry,

^{101}C_r = ^{101}C_{101-r}

Hence, the sum from $r = 51$ to $r = 101$ is exactly half of the total sum. Therefore,

^{101}C_{51} + ^{101}C_{52} + \dots + ^{101}C_{101} = \frac{2^{101}}{2} = 2^{100}

So,

S = \frac{2^{100}}{101}

Therefore, the correct option is D.

Using symmetry of binomial coefficients

Given:

\frac{^{100}C_{50}}{51} + \frac{^{100}C_{51}}{52} + \dots + \frac{^{100}C_{100}}{101}

Find: Its numerical value.

First convert each term using

\frac{^{100}C_r}{r+1} = \frac{^{101}C_{r+1}}{101}

Thus,

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

Let $k = r+1$ . Then $k$ runs from $51$ to $101$ , so

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

For odd upper index $101$ , the coefficients split into two equal halves because

^{101}C_k = ^{101}C_{101-k}

and there is no middle unpaired term. Hence,

\sum_{k=51}^{101} {}^{101}C_k = \frac{1}{2}\sum_{k=0}^{101} {}^{101}C_k = \frac{1}{2} \cdot 2^{101} = 2^{100}

Therefore,

S = \frac{1}{101} \cdot 2^{100} = \frac{2^{100}}{101}

Therefore, the value is $\frac{2^{100}}{101}$ and the correct option is D.

Using $\frac{^{n}C_r}{r+1}$ incorrectly is a common mistake. The correct identity is $\frac{^{n}C_r}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}$ . If the index shift is missed, the transformed sum becomes wrong.
Taking the sum from $51$ to $101$ as more than half of the total is incorrect. For odd upper index $101$ , symmetry pairs every term exactly, so the coefficients from $51$ to $101$ form exactly one half of the total sum.
Forgetting to divide by $101$ after converting the terms is wrong. The factor $\frac{1}{101}$ comes from the identity and must remain outside the entire sum.

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