If `"""""^(2n)C_3`` : "^nC_3 =10 :1`, then the ratio `(n^2+3n) : (n^2 - 3n+4)` is

Options

A. `35 : 16`
B. ` 27 : 11`
C. `65 : 37`
D. `2 : 1`

Answer

To solve the problem, we need to find the value of \( n \) given the ratio of binomial coefficients and then use that value to find the ratio of two expressions involving \( n \).

### Step 1: Set up the equation from the given ratio
We start with the equation given in the problem:

\[ \frac{{^{2n}C_3}}{{^{n}C_3}} = \frac{10}{1} \]

### Step 2: Write the binomial coefficient expressions
Using the formula for binomial coefficients, we can express \( ^{2n}C_3 \) and \( ^{n}C_3 \):

\[ ^{2n}C_3 = \frac{{(2n)!}}{{3!(2n-3)!}} \quad \text{and} \quad ^{n}C_3 = \frac{{n!}}{{3!(n-3)!}} \]

### Step 3: Substitute into the ratio
Substituting these into the ratio gives:

\[ \frac{{\frac{{(2n)!}}{{3!(2n-3)!}}}}{{\frac{{n!}}{{3!(n-3)!}}}} = 10 \]

### Step 4: Simplify the equation
The \( 3! \) cancels out:

\[ \frac{{(2n)!}}{{(2n-3)!}} \cdot \frac{{(n-3)!}}{{n!}} = 10 \]

### Step 5: Expand the factorials
We can expand \( (2n)! \) and \( (2n-3)! \):

\[ (2n)(2n-1)(2n-2) \cdot \frac{{(n-3)!}}{{n!}} = 10 \]

### Step 6: Simplify further
Now, we can express \( \frac{{(n-3)!}}{{n!}} \) as \( \frac{1}{{n(n-1)(n-2)}} \):

\[ (2n)(2n-1)(2n-2) \cdot \frac{1}{{n(n-1)(n-2)}} = 10 \]

### Step 7: Cross-multiply to eliminate the fraction
Cross-multiplying gives:

\[ (2n)(2n-1)(2n-2) = 10n(n-1)(n-2) \]

### Step 8: Expand both sides
Expanding both sides leads to:

\[ 8n^3 - 12n^2 + 4n = 10n^3 - 30n^2 + 20n \]

### Step 9: Rearrange the equation
Rearranging gives:

\[ 0 = 2n^3 - 18n^2 + 16n \]

### Step 10: Factor out common terms
Factoring out \( 2n \):

\[ 2n(n^2 - 9n + 8) = 0 \]

### Step 11: Solve the quadratic equation
Setting \( n^2 - 9n + 8 = 0 \):

Using the quadratic formula:

\[ n = \frac{{9 \pm \sqrt{{81 - 32}}}}{2} = \frac{{9 \pm \sqrt{49}}}{2} = \frac{{9 \pm 7}}{2} \]

This gives us:

\[ n = 8 \quad \text{or} \quad n = 1 \]

### Step 12: Choose the appropriate value of \( n \)
Since \( n \) must be greater than or equal to 3 for \( ^{n}C_3 \) to be defined, we take \( n = 8 \).

### Step 13: Calculate the required ratio
Now, we need to find the ratio \( (n^2 + 3n) : (n^2 - 3n + 4) \):

Substituting \( n = 8 \):

\[ n^2 + 3n = 64 + 24 = 88 \]
\[ n^2 - 3n + 4 = 64 - 24 + 4 = 44 \]

Thus, the ratio is:

\[ \frac{88}{44} = 2 : 1 \]

### Final Answer
The ratio \( (n^2 + 3n) : (n^2 - 3n + 4) \) is \( 2 : 1 \).

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