`1 + ""^2C_1x + ""^3C_2 x^2+ ""^4C_3 x^3 + …..` to ` oo` terms can be summed up if

Options

A. a) `x lt 1`
B. b) `x gt - 1`
C. c) `-1 lt x lt 1`
D. d) `-oo lt x lt oo`

Answer

To solve the problem, we need to analyze the series given:

\[ S = 1 + \binom{2}{1} x + \binom{3}{2} x^2 + \binom{4}{3} x^3 + \ldots \]

### Step 1: Rewrite the Binomial Coefficients
We can express the binomial coefficients using the formula for combinations:

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

Applying this to our series, we have:

\[ S = 1 + \frac{2!}{1!(2-1)!} x + \frac{3!}{2!(3-2)!} x^2 + \frac{4!}{3!(4-3)!} x^3 + \ldots \]

### Step 2: Simplify Each Term
Now, let's simplify each term:

- The first term is \( 1 \).
- The second term is \( 2x \).
- The third term is \( \frac{3 \cdot 2}{2} x^2 = 3x^2 \).
- The fourth term is \( \frac{4 \cdot 3 \cdot 2}{6} x^3 = 4x^3 \).

Thus, we can rewrite \( S \) as:

\[ S = 1 + 2x + 3x^2 + 4x^3 + \ldots \]

### Step 3: Identify the Pattern
Notice that the coefficients \( 1, 2, 3, 4, \ldots \) can be expressed as \( n \) where \( n \) is the index of the term. Therefore, we can express \( S \) in a more general form:

\[ S = \sum_{n=0}^{\infty} (n+1)x^n \]

### Step 4: Use the Formula for the Sum of Series
We can use the formula for the sum of an infinite series:

\[ \sum_{n=0}^{\infty} nx^n = \frac{x}{(1-x)^2} \]

Thus, we can derive:

\[ \sum_{n=0}^{\infty} (n+1)x^n = \sum_{n=0}^{\infty} nx^n + \sum_{n=0}^{\infty} x^n = \frac{x}{(1-x)^2} + \frac{1}{1-x} \]

### Step 5: Combine the Results
Now we combine the results:

\[ S = \frac{x}{(1-x)^2} + \frac{1}{1-x} \]

### Step 6: Simplify the Expression
To combine these fractions, we find a common denominator:

\[ S = \frac{x + (1-x)(1-x)}{(1-x)^2} = \frac{x + (1 - 2x + x^2)}{(1-x)^2} = \frac{1 - x + x^2}{(1-x)^2} \]

### Step 7: Determine the Convergence
The series converges if the absolute value of \( x \) is less than 1:

\[ |x| < 1 \]

### Final Answer
Thus, the series \( S = 1 + \binom{2}{1} x + \binom{3}{2} x^2 + \binom{4}{3} x^3 + \ldots \) can be summed up if:

\[ |x| < 1 \]