The value of
`((C_(1))/(C_(0))+2(C_(2))/(C_(1))+3(C_(3))/(C_(2))+"......"+10(C_(10))/(C_(9)))`
(where `C_(r) = .^(10)C_(r)`), is `11lambda`, then value of `lambda` is

Options

A. `1`
B. `2`
C. `4`
D. `5`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \frac{C_1}{C_0} + 2 \cdot \frac{C_2}{C_1} + 3 \cdot \frac{C_3}{C_2} + \ldots + 10 \cdot \frac{C_{10}}{C_9} \]

where \( C_r = \binom{10}{r} \).

### Step 1: Rewrite the expression using the definition of combinations

We know that:

\[ C_r = \binom{10}{r} = \frac{10!}{r!(10-r)!} \]

Thus, we can express the ratios as follows:

\[ \frac{C_r}{C_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10!}{r!(10-r)!} \cdot \frac{(r-1)!(10-(r-1))!}{10!} = \frac{10 - (r-1)}{r} = \frac{10 - r + 1}{r} \]

### Step 2: Substitute the ratios back into the expression

Now we can rewrite the entire expression:

\[ \sum_{r=1}^{10} r \cdot \frac{C_r}{C_{r-1}} = \sum_{r=1}^{10} r \cdot \frac{10 - r + 1}{r} = \sum_{r=1}^{10} (10 - r + 1) \]

### Step 3: Simplify the sum

This simplifies to:

\[ \sum_{r=1}^{10} (11 - r) \]

### Step 4: Calculate the sum

Now we can calculate the sum:

\[ \sum_{r=1}^{10} (11 - r) = 11 \cdot 10 - \sum_{r=1}^{10} r = 110 - \frac{10 \cdot 11}{2} = 110 - 55 = 55 \]

### Step 5: Set the sum equal to \( 11\lambda \)

According to the problem, this sum equals \( 11\lambda \):

\[ 55 = 11\lambda \]

### Step 6: Solve for \( \lambda \)

Now we can solve for \( \lambda \):

\[ \lambda = \frac{55}{11} = 5 \]

Thus, the value of \( \lambda \) is:

\[ \boxed{5} \]

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