Let `f(x) = 1 - x +x^2-x^3+......+x^16+x^17 , then coefficient of x^2` in f(x-1) is?

Answer

To find the coefficient of \( x^2 \) in \( f(x-1) \), where

\[ f(x) = 1 - x + x^2 - x^3 + \ldots + x^{17}, \]

we can first express \( f(x) \) in a more manageable form. The series is a geometric series with the first term \( a = 1 \) and common ratio \( r = -x \). The number of terms in the series is \( 18 \) (from \( x^0 \) to \( x^{17} \)).

### Step 1: Write the formula for the geometric series

The sum of a geometric series can be expressed as:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

where \( n \) is the number of terms. In our case, we have:

\[ f(x) = \frac{1 - (-x)^{18}}{1 - (-x)} = \frac{1 - x^{18}}{1 + x}. \]

### Step 2: Substitute \( x - 1 \) into \( f(x) \)

Now we need to find \( f(x-1) \):

\[ f(x-1) = \frac{1 - (x-1)^{18}}{1 + (x-1)} = \frac{1 - (x-1)^{18}}{x}. \]

### Step 3: Expand \( (x-1)^{18} \) using the Binomial Theorem

Using the Binomial Theorem, we can expand \( (x-1)^{18} \):

\[ (x-1)^{18} = \sum_{k=0}^{18} \binom{18}{k} x^{18-k} (-1)^k. \]

### Step 4: Simplify \( f(x-1) \)

Substituting this expansion back into \( f(x-1) \):

\[ f(x-1) = \frac{1 - \sum_{k=0}^{18} \binom{18}{k} x^{18-k} (-1)^k}{x}. \]

This simplifies to:

\[ f(x-1) = \frac{1 - (x^{18} - 18x^{17} + \ldots + (-1)^{18})}{x}. \]

### Step 5: Find the coefficient of \( x^2 \)

To find the coefficient of \( x^2 \) in \( f(x-1) \), we need to consider the terms in the numerator that will contribute to \( x^3 \) after division by \( x \):

1. The constant term \( 1 \) contributes \( 0 \) to \( x^2 \).
2. The term \( -(-1)^{18} \) contributes \( 1 \) (the constant term).
3. The term \( -\binom{18}{2} x^{16} \) contributes \( -\binom{18}{2} \) to \( x^2 \).

Calculating \( \binom{18}{2} \):

\[ \binom{18}{2} = \frac{18 \times 17}{2} = 153. \]

Thus, the coefficient of \( x^2 \) in \( f(x-1) \) is:

\[ -153 + 1 = -152. \]

### Final Answer

The coefficient of \( x^2 \) in \( f(x-1) \) is \( -152 \).

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