If y=`(1 +x)^n ` then the value of `(y_0)+(y_1)_0 +((y_2)_0)/(2!) +((y_3)_0)/(3!) +. .. ((y_n)_0)/(n!)` is

Options

A. `2^n`
B. `2^(n-1)`
C. `n`
D. None

Answer

To solve the problem step by step, we start with the given function:

**Step 1: Identify the function**
Let \( y = (1 + x)^n \).

**Step 2: Calculate \( y_0 \)**
To find \( y_0 \), we substitute \( x = 0 \):
\[ y_0 = (1 + 0)^n = 1^n = 1. \]

**Step 3: Calculate \( y_1 \)**
Now, we need to find the first derivative \( y' \) and evaluate it at \( x = 0 \):
\[ y' = \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1}. \]
Evaluating at \( x = 0 \):
\[ y_1 = y'(0) = n(1 + 0)^{n-1} = n. \]

**Step 4: Calculate \( y_2 \)**
Next, we find the second derivative \( y'' \) and evaluate it at \( x = 0 \):
\[ y'' = \frac{d^2}{dx^2}[(1 + x)^n] = n(n-1)(1 + x)^{n-2}. \]
Evaluating at \( x = 0 \):
\[ y_2 = y''(0) = n(n-1)(1 + 0)^{n-2} = n(n-1). \]

**Step 5: Calculate \( y_3 \)**
Now, we find the third derivative \( y''' \) and evaluate it at \( x = 0 \):
\[ y''' = \frac{d^3}{dx^3}[(1 + x)^n] = n(n-1)(n-2)(1 + x)^{n-3}. \]
Evaluating at \( x = 0 \):
\[ y_3 = y'''(0) = n(n-1)(n-2)(1 + 0)^{n-3} = n(n-1)(n-2). \]

**Step 6: Generalize \( y_k \)**
From the pattern observed, we can generalize:
\[ y_k = n(n-1)(n-2)\cdots(n-k+1) = \frac{n!}{(n-k)!}. \]

**Step 7: Formulate the series**
Now we can write the series:
\[ S = y_0 + \frac{y_1}{1!} + \frac{y_2}{2!} + \frac{y_3}{3!} + \cdots + \frac{y_n}{n!}. \]

Substituting the values we found:
\[ S = 1 + \frac{n}{1!} + \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} + \cdots + \frac{n!}{n!}. \]

**Step 8: Recognize the binomial expansion**
This series represents the expansion of \( (1 + 1)^n \):
\[ S = (1 + 1)^n = 2^n. \]

**Final Result:**
Thus, the value of the series is:
\[ \boxed{2^n}. \]