Let `S_n=1+q+q^2 +...+q^n` and `T_n =1+((q+1)/2)+((q+1)/2)^2+...((q+1)/2)^n` If `alpha T_100=^101C_1 +^101C_2` x `S_1 ...+^101C_101` x`S_100,` then the value of `alpha` is equal to (A) `2^99` (B) `2^101` (C) `2^100` (D) `-2^100`

Options

A. `2^(100)`
B. 200
C. `2^(99)`
D. 202

Answer

To solve the problem, we need to find the value of \( \alpha \) given the expressions for \( S_n \) and \( T_n \) and the equation involving binomial coefficients.

### Step 1: Define \( S_n \) and \( T_n \)

The expression for \( S_n \) is given as:
\[ S_n = 1 + q + q^2 + \ldots + q^n \]
This is a geometric series with first term \( a = 1 \) and common ratio \( r = q \). The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^{n+1} - 1)}{r - 1} = \frac{1(q^{n+1} - 1)}{q - 1} = \frac{q^{n+1} - 1}{q - 1} \]

The expression for \( T_n \) is given as:
\[ T_n = 1 + \left(\frac{q + 1}{2}\right) + \left(\frac{q + 1}{2}\right)^2 + \ldots + \left(\frac{q + 1}{2}\right)^n \]
This is also a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{q + 1}{2} \). The sum is:
\[ T_n = \frac{1\left(\left(\frac{q + 1}{2}\right)^{n+1} - 1\right)}{\frac{q + 1}{2} - 1} = \frac{\left(\frac{q + 1}{2}\right)^{n+1} - 1}{\frac{q - 1}{2}} = \frac{2\left(\left(\frac{q + 1}{2}\right)^{n+1} - 1\right)}{q - 1} \]

### Step 2: Evaluate \( T_{100} \)

Substituting \( n = 100 \) into the expression for \( T_n \):
\[ T_{100} = \frac{2\left(\left(\frac{q + 1}{2}\right)^{101} - 1\right)}{q - 1} \]

### Step 3: Set up the equation involving \( \alpha \)

The problem states:
\[ \alpha T_{100} = \binom{101}{1} S_1 + \binom{101}{2} S_2 + \ldots + \binom{101}{101} S_{100} \]

We can express the right-hand side in summation form:
\[ \alpha T_{100} = \sum_{r=1}^{101} \binom{101}{r} S_{r-1} \]

### Step 4: Substitute \( S_{r-1} \)

Using the formula for \( S_n \):
\[ S_{r-1} = \frac{q^r - 1}{q - 1} \]

Thus, the right-hand side becomes:
\[ \alpha T_{100} = \sum_{r=1}^{101} \binom{101}{r} \frac{q^r - 1}{q - 1} \]

This can be separated into two sums:
\[ \alpha T_{100} = \frac{1}{q - 1} \left( \sum_{r=1}^{101} \binom{101}{r} q^r - \sum_{r=1}^{101} \binom{101}{r} \right) \]

The second sum simplifies to \( 2^{101} - 1 \) (the sum of all binomial coefficients).

### Step 5: Evaluate the first sum

Using the binomial theorem:
\[ \sum_{r=0}^{101} \binom{101}{r} q^r = (1 + q)^{101} \]
Thus,
\[ \sum_{r=1}^{101} \binom{101}{r} q^r = (1 + q)^{101} - 1 \]

### Step 6: Substitute back into the equation

Now we have:
\[ \alpha T_{100} = \frac{1}{q - 1} \left( (1 + q)^{101} - 1 - (2^{101} - 1) \right) \]
Simplifying gives:
\[ \alpha T_{100} = \frac{(1 + q)^{101} - 2^{101}}{q - 1} \]

### Step 7: Solve for \( \alpha \)

We know \( T_{100} \) from earlier:
\[ T_{100} = \frac{2\left(\left(\frac{q + 1}{2}\right)^{101} - 1\right)}{q - 1} \]
Substituting \( T_{100} \) into the equation:
\[ \alpha \cdot \frac{2\left(\left(\frac{q + 1}{2}\right)^{101} - 1\right)}{q - 1} = \frac{(1 + q)^{101} - 2^{101}}{q - 1} \]
Cancelling \( \frac{1}{q - 1} \) gives:
\[ \alpha \cdot 2\left(\frac{(q + 1)^{101}}{2^{101}} - 1\right) = (1 + q)^{101} - 2^{101} \]

### Step 8: Final Calculation

Solving for \( \alpha \) gives:
\[ \alpha = 2^{100} \]

Thus, the value of \( \alpha \) is:
\[ \boxed{2^{100}} \]