Question

If $^{100}C_0 + ^{101}C_1 + ^{102}C_2 + ^{103}C_3 + \dots + ^{200}C_{100} = ^nC_r$, then:

• A. Least value of $n+r$ is 301
• B. Least value of $n+r$ is 302
• C. $n$ is a prime number
• D. $r$ can be a prime number

Answer 1

Answer: (A), (D)

Least value of $n+r$ is 301, $r$ can be a prime number

Answer 2

We use the identity:

$$\sum_{k=0}^{m}\binom{n+k}{k}=\binom{n+m+1}{m}$$

Here, take $n=100$ and $m=100$:

$$^{100}C_0 + ^{101}C_1 + ^{102}C_2 + \dots + ^{200}C_{100} = \binom{100+100+1}{100} = \binom{201}{100}.$$

Thus, we have $n = 201$ and $r = 100$, so:

$$n + r = 201 + 100 = 301.$$

Also, note that by the symmetry property of binomial coefficients, we have:

$$\binom{201}{100} = \binom{201}{201-100} = \binom{201}{101},$$

and since $101$ is a prime number, we see that indeed $r$ can be a prime number.