The value of \; {{}¹00}C₅0}}{51} + \; {{}¹00}C₅1}}{52} + + \; {{}¹00}C₁00}}{101} is: [23 Jan 2026 Shift 1]

Correct Answer is : 2100101\; {2¹00}}{101}1012100

Correct Answer is : 2100101\; {2¹00}}{101}1012100

Test Index

Binomial Theorem Part 3

Section: Mathematics

Question : 13 of 53

Marks: +1, -0

The value of

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

Solution:

Concept:

Use the identity

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

to convert the sum into a binomial sum.

Explanation:

Write the given sum as

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

Apply the identity:

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

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

Change index: let

k = r+1

. When

r=50

k=51

; when

r=100

k=101

Thus

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

Total sum of all binomial coefficients:

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

By symmetry,

\sum_{k=0}^{50} {}^{101}C_k = \sum_{k=51}^{101} {}^{101}C_k = 2^{100}

Therefore

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

Answer:

Option B:

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

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