If `.^25 C_0 + 5 .^25 C_1 + 9 .^25 C_2 .... 101 .^25 C_25 = 2^(25) k` find k = ?

Answer

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

\[ S = \sum_{r=0}^{25} (4r + 1) \cdot \binom{25}{r} \]

This can be split into two separate sums:

\[ S = \sum_{r=0}^{25} 4r \cdot \binom{25}{r} + \sum_{r=0}^{25} \binom{25}{r} \]

### Step 1: Evaluate the first sum \(\sum_{r=0}^{25} 4r \cdot \binom{25}{r}\)

Using the identity for the sum of \(r \cdot \binom{n}{r}\):

\[ \sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1} \]

For \(n = 25\):

\[ \sum_{r=0}^{25} r \cdot \binom{25}{r} = 25 \cdot 2^{24} \]

Thus,

\[ \sum_{r=0}^{25} 4r \cdot \binom{25}{r} = 4 \cdot 25 \cdot 2^{24} = 100 \cdot 2^{24} \]

### Step 2: Evaluate the second sum \(\sum_{r=0}^{25} \binom{25}{r}\)

Using the binomial theorem:

\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

For \(n = 25\):

\[ \sum_{r=0}^{25} \binom{25}{r} = 2^{25} \]

### Step 3: Combine the results

Now we can combine the two sums:

\[ S = 100 \cdot 2^{24} + 2^{25} \]

Factoring out \(2^{24}\):

\[ S = 2^{24} (100 + 2) = 2^{24} \cdot 102 \]

### Step 4: Express \(S\) in the required form

We need to express \(S\) in the form \(2^{25} k\):

\[ S = 2^{24} \cdot 102 = 2^{25} \cdot \frac{102}{2} = 2^{25} \cdot 51 \]

### Conclusion

Thus, we find that:

\[ k = 51 \]