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Mathematics · Permutations and Combinations

JEE Main 2026 — 23 January, Morning Shift — Question 12

The value of 100C5051+100C5152+….+100C100101\frac{{ }^{100} \mathrm{C}_{50}}{51}+\frac{{ }^{100} \mathrm{C}_{51}}{52}+\ldots .+\frac{{ }^{100} \mathrm{C}_{100}}{101} is :

  1. Option A:

    2101100\frac{2^{101}}{100}

  2. Option B:

    2100100\frac{2^{100}}{100}

  3. Option C:

    2101101\frac{2^{101}}{101}

  4. Option D:

    2100101\frac{2^{100}}{101}

    Correct

Answer: D

Step-by-step solution

S=∑r=50100100Crr+1=∑r=501001r+1⋅r+1101⋅101Cr+1\mathrm{S}=\sum_{\mathrm{r}=50}^{100} \frac{{ }^{100} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}=\sum_{\mathrm{r}=50}^{100} \frac{1}{\mathrm{r}+1} \cdot \frac{\mathrm{r}+1}{101} \cdot{ }^{101} \mathrm{C}_{\mathrm{r}+1}

S=1101∑r=50100101Cr+1\mathrm{S}=\frac{1}{101} \sum_{\mathrm{r}=50}^{100}{ }^{101} \mathrm{C}_{\mathrm{r}+1}

=1101×21012=2100101=\frac{1}{101} \times \frac{2^{101}}{2}=\frac{2^{100}}{101}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Combinations

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