Question

If \(^nC_r = C_r\) and \[ 2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650 \] then \[ ^nC_2 = \ ? \]

• A. \(25\)
• B. \(300\)
• C. \(225\)
• D. \(625\)

Hint:

For combinatorial identities and sums, leverage known binomial coefficient properties and identities, such as the binomial theorem or Pascal's triangle relations.

Answer 1

The Correct Option is B

We are given that
\[ 2\frac{C_1}{C_0} + 4\frac{C_2}{C_1} + 6\frac{C_3}{C_2} + \dots + 2n\frac{C_n}{C_{n-1}} = 650 \] We know that \(\frac{C_r}{C_{r-1}} = \frac{{^n}C_r}{{^n}C_{r-1}} = \frac{n-r+1}{r}\).
So, the given expression can be written as:
\[\sum_{r=1}^n 2r \frac{C_r}{C_{r-1}} = \sum_{r=1}^n 2r \frac{n-r+1}{r} = 650\]
\[\sum_{r=1}^n 2(n-r+1) = 650\]
\[2\sum_{r=1}^n (n-r+1) = 650\]
\[\sum_{r=1}^n (n-r+1) = 325\]
Let \(S = \sum_{r=1}^n (n-r+1)\).
When \(r=1\), the term is \(n\).
When \(r=2\), the term is \(n-1\).
When \(r=3\), the term is \(n-2\).
And so on, until \(r=n\), the term is \(1\).
So, \(S = n + (n-1) + (n-2) + \dots + 1\).
This is the sum of the first \(n\) natural numbers, which is \(\frac{n(n+1)}{2}\).
Therefore,
\[\frac{n(n+1)}{2} = 325\] \[n(n+1) = 650\] \[n^2 + n - 650 = 0\] We need to find two numbers whose product is -650 and whose sum is 1.
We can write \(650 = 25 \times 26\).
So, \(n^2 + 26n - 25n - 650 = 0\)
\(n(n+26) - 25(n+26) = 0\)
\((n-25)(n+26) = 0\)
Since \(n\) must be positive, \(n=25\).
Now we need to find \({^n}C_2 = {^{25}}C_2\).
\[{^{25}}C_2 = \frac{25 \times 24}{2} = 25 \times 12 = 300\] Therefore, \({^n}C_2 = 300\). Final Answer: The final answer is $\boxed{(2)}$