Let `(1 + x)^(10)=sum_(r=0)^(10)c_(r)x^(r)` and `(1+x)^(7)=sum_(r=0)^(7)d_(r),x^(r).` If `P=sum_(r=0)^(5)c_(2r)` and `Q=sum_(r=0)^(3)d_(2r+1)`, then `P/Q` is equal to ______

Answer

To solve the problem, we need to calculate the values of \( P \) and \( Q \) based on the given binomial expansions and then find \( \frac{P}{Q} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Expansions:**
We have:
\[ (1 + x)^{10} = \sum_{r=0}^{10} c_r x^r \]
and
\[ (1 + x)^{7} = \sum_{r=0}^{7} d_r x^r \]
where \( c_r = \binom{10}{r} \) and \( d_r = \binom{7}{r} \).

2. **Calculating \( P \):**
\( P \) is defined as:
\[ P = \sum_{r=0}^{5} c_{2r} \]
This means we need to sum the coefficients of the even powers of \( x \) in the expansion of \( (1 + x)^{10} \):
\[ P = c_0 + c_2 + c_4 + c_6 + c_8 + c_{10} \]
Using the property of binomial coefficients, we can express this as:
\[ P = \frac{(1 + 1)^{10} + (1 - 1)^{10}}{2} = \frac{2^{10} + 0}{2} = \frac{1024}{2} = 512 \]

3. **Calculating \( Q \):**
\( Q \) is defined as:
\[ Q = \sum_{r=0}^{3} d_{2r+1} \]
This means we need to sum the coefficients of the odd powers of \( x \) in the expansion of \( (1 + x)^{7} \):
\[ Q = d_1 + d_3 + d_5 + d_7 \]
Again, using the property of binomial coefficients:
\[ Q = \frac{(1 + 1)^{7} - (1 - 1)^{7}}{2} = \frac{2^{7} - 0}{2} = \frac{128}{2} = 64 \]

4. **Finding \( \frac{P}{Q} \):**
Now we can find \( \frac{P}{Q} \):
\[ \frac{P}{Q} = \frac{512}{64} = 8 \]

Thus, the final answer is:
\[ \frac{P}{Q} = 8 \]