Binomial Theorem – Summation of Series – JEE Main 23 Jan 2026 Shift 1

Solution:

Let the sum be $S$.
$$ S = \sum_{r=50}^{100} \frac{^{100}C_r}{r+1} $$

Using the property $\frac{^nC_r}{r+1} = \frac{1}{n+1} \cdot ^{n+1}C_{r+1}$:
$$ S = \sum_{r=50}^{100} \frac{1}{101} \cdot ^{101}C_{r+1} $$

Let $k = r+1$. When $r=50, k=51$. When $r=100, k=101$.
$$ S = \frac{1}{101} \sum_{k=51}^{101} (^{101}C_k) $$
$$ S = \frac{1}{101} \left( ^{101}C_{51} + ^{101}C_{52} + \dots + ^{101}C_{101} \right) $$

We know that the sum of binomial coefficients is $\sum_{k=0}^{n} (^nC_k) = 2^n$.
Here $n=101$ (an odd number).
For odd $n$, the sum of the first half of coefficients equals the sum of the second half:
$$ \sum_{k=0}^{50} (^{101}C_k) = \sum_{k=51}^{101} (^{101}C_k) = \frac{2^{101}}{2} = 2^{100} $$

Therefore:
$$ S = \frac{1}{101} \cdot (2^{100}) $$
$$ S = \frac{2^{100}}{101} $$

Ans. (4)