Let q be a positive with `q le 50`.
If the sum `""^(98)C_(30)+2" "^(97)C_(30)+3." "^(96)C_(30)+ …… + 68." "^(31)C_(30)+69." "^(30)C_(30)=""^(100)C_(q)`
Find the sum of the digits of q.

Answer

To solve the given problem, we need to evaluate the sum

\[ \sum_{j=30}^{69} (70-j) \binom{98-j}{30} \]

and show that it is equal to \(\binom{100}{q}\) for some \(q\) where \(q \leq 50\).

### Step 1: Understanding the sum

The given sum can be rewritten as:

\[ \sum_{j=30}^{69} (70-j) \binom{98-j}{30} \]

This can be interpreted as a weighted sum of binomial coefficients.

### Step 2: Applying the formula

We can use the identity for binomial coefficients:

\[ \sum_{j=r}^{n} (n-j+1) \binom{j}{r} = \binom{n+2}{r+2} \]

In our case, we can set \(n = 98\) and \(r = 30\):

\[ \sum_{j=30}^{98} (99-j) \binom{j}{30} = \binom{100}{32} \]

### Step 3: Adjusting the limits

However, our original sum only goes from \(j=30\) to \(j=69\). We can adjust our sum accordingly:

\[ \sum_{j=30}^{69} (70-j) \binom{98-j}{30} = \sum_{j=30}^{98} (70-j) \binom{98-j}{30} - \sum_{j=70}^{98} (70-j) \binom{98-j}{30} \]

The first part can be computed using the identity we mentioned, while the second part can be computed separately.

### Step 4: Simplifying the sum

Using the identity, we have:

\[ \sum_{j=30}^{98} (70-j) \binom{98-j}{30} = \binom{100}{32} \]

Now, we need to find the value of \(q\) such that:

\[ \binom{100}{q} = \binom{100}{32} \]

### Step 5: Finding \(q\)

From the properties of binomial coefficients, we know:

\[ \binom{n}{k} = \binom{n}{n-k} \]

Thus, \(q\) can be either \(32\) or \(100 - 32 = 68\). Since \(q\) must be less than or equal to \(50\), we take:

\[ q = 32 \]

### Step 6: Finding the sum of the digits of \(q\)

Now, we need to find the sum of the digits of \(q\):

\[ 3 + 2 = 5 \]

### Conclusion

Thus, the final answer is:

\[ \text{The sum of the digits of } q \text{ is } 5. \]

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