Question

If $\frac{{}^9C_0}{8} - \frac{{}^9C_1}{9} + \frac{{}^9C_2}{10} - \frac{{}^9C_3}{11} + ... + \frac{{}^9C_8}{16} - \frac{{}^9C_9}{17} = \frac{1}{n}$ then n is divisible by

• A. 7
• B. 11
• C. 13
• D. 17

Answer 1

Answer: (B), (C), (D)

11, 13, 17

Answer 2

The given series is $S = \frac{{}^9C_0}{8} - \frac{{}^9C_1}{9} + \frac{{}^9C_2}{10} - \frac{{}^9C_3}{11} + ... + \frac{{}^9C_8}{16} - \frac{{}^9C_9}{17}$. This can be written in summation form as $S = \sum_{r=0}^{9} (-1)^r \frac{{}^9C_r}{8+r}$.

We use the integral identity for binomial coefficients: $\sum_{r=0}^{n} (-1)^r \frac{{}^nC_r}{k+r} = \int_0^1 x^{k-1} (1-x)^n dx$.

In our case, $n=9$ and $k=8$. So, the sum $S$ can be expressed as: $S = \int_0^1 x^{8-1} (1-x)^9 dx = \int_0^1 x^7 (1-x)^9 dx$.

This integral is a Beta function, $B(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1} dx = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. Here, $a-1=7 \Rightarrow a=8$ and $b-1=9 \Rightarrow b=10$. So, $S = B(8, 10) = \frac{\Gamma(8)\Gamma(10)}{\Gamma(8+10)} = \frac{\Gamma(8)\Gamma(10)}{\Gamma(18)}$.

Using the property $\Gamma(m) = (m-1)!$ for positive integers $m$: $S = \frac{(8-1)!(10-1)!}{(18-1)!} = \frac{7! 9!}{17!}$.

Now, we need to simplify this expression: $S = \frac{7! \times 9!}{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9!}$. Cancel $9!$ from numerator and denominator: $S = \frac{7!}{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}$.

We know $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$. The denominator is $D = 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10$.

Let's simplify by cancelling common factors: $D = 17 \times (2^4) \times (3 \times 5) \times (2 \times 7) \times 13 \times (2^2 \times 3) \times 11 \times (2 \times 5)$. $D = 17 \times 13 \times 11 \times (2^{4+1+2+1}) \times (3^{1+1}) \times (5^{1+1}) \times 7^1$. $D = 17 \times 13 \times 11 \times 2^8 \times 3^2 \times 5^2 \times 7$.

Numerator $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7 \times (2 \times 3) \times 5 \times (2^2) \times 3 \times 2 = 2^4 \times 3^2 \times 5 \times 7$.

Now, substitute these prime factorizations into the expression for $S$: $S = \frac{2^4 \times 3^2 \times 5 \times 7}{17 \times 13 \times 11 \times 2^8 \times 3^2 \times 5^2 \times 7}$.

Cancel common factors from numerator and denominator: $2^4$ in numerator cancels with $2^8$ in denominator, leaving $2^4$ in denominator. $3^2$ cancels out completely. $5$ in numerator cancels with $5^2$ in denominator, leaving $5$ in denominator. $7$ cancels out completely.

So, $S = \frac{1}{17 \times 13 \times 11 \times 2^4 \times 5}$. $S = \frac{1}{17 \times 13 \times 11 \times 16 \times 5}$.

Calculate the value of the denominator: $16 \times 5 = 80$. $11 \times 13 = 143$. $17 \times 80 = 1360$. $1360 \times 143 = 194480$.

So, $S = \frac{1}{194480}$. The problem states that $S = \frac{1}{n}$, so $n = 194480$.

We need to find which of the given options (7, 11, 13, 17) divides $n$. From the prime factorization of $n$ (which is the denominator calculated above): $n = 17 \times 13 \times 11 \times 2^4 \times 5$. The prime factors of $n$ are $2, 5, 11, 13, 17$.

Checking the options:

(A) 7: $n$ is not divisible by 7. (B) 11: $n$ is divisible by 11. (C) 13: $n$ is divisible by 13. (D) 17: $n$ is divisible by 17.

Since the question asks "then n is divisible by", and multiple options are correct, we must select all correct options.