The value of `""^(40) C_(31) + sum _(r = 0)^(10) ""^(40 + r) C_(10 +r)` is equal to

Options

A. `""^(51)C_(20)`
B. `2""^(50)C_(20)`
C. `2""^(45)C_(15)`
D. `""^(51)C_(21)`

Answer

To solve the problem \( {}^{40}C_{31} + \sum_{r=0}^{10} {}^{40+r}C_{10+r} \), we will break it down step by step.

### Step 1: Rewrite the Summation
The summation can be rewritten as:
\[ \sum_{r=0}^{10} {}^{40+r}C_{10+r} = {}^{40}C_{10} + {}^{41}C_{11} + {}^{42}C_{12} + \ldots + {}^{50}C_{20} \]

### Step 2: Combine the Terms
Now, we can combine the first term \( {}^{40}C_{31} \) with the rewritten summation:
\[ {}^{40}C_{31} + {}^{40}C_{10} + {}^{41}C_{11} + {}^{42}C_{12} + \ldots + {}^{50}C_{20} \]

### Step 3: Apply Pascal's Identity
Using Pascal's identity, which states that \( {}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1} \), we can simplify the expression:
- First, we can combine \( {}^{40}C_{31} \) and \( {}^{40}C_{10} \):
\[ {}^{40}C_{31} + {}^{40}C_{10} = {}^{41}C_{11} \]

### Step 4: Continue Applying Pascal's Identity
Now, we can continue applying Pascal's identity to the remaining terms:
\[ {}^{41}C_{11} + {}^{41}C_{11} = {}^{42}C_{12} \]
\[ {}^{42}C_{12} + {}^{42}C_{12} = {}^{43}C_{13} \]
Continuing this process, we eventually combine all the way up to:
\[ {}^{50}C_{20} \]

### Step 5: Final Calculation
After applying Pascal's identity repeatedly, we find that:
\[ {}^{40}C_{31} + {}^{40}C_{10} + {}^{41}C_{11} + \ldots + {}^{50}C_{20} = {}^{51}C_{20} \]

### Conclusion
Thus, the final answer is:
\[ \text{The value is } {}^{51}C_{20} \]

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