Question

Using the identity $\frac{1}{^{2n+1}C_r} + \frac{1}{^{2n+1}C_{r+1}} = \frac{2n+2}{2n+1} \frac{1}{^{2n}C_r}$, then value of $\sum_{r=1}^{19} \frac{(-1)^r r}{^{20}C_r} = -\frac{p}{q}$, where p and q are relatively prime natural numbers, then p + q is equal to

Answer 1

21

Answer 2

The problem asks us to evaluate the sum $\sum_{r=1}^{19} \frac{(-1)^r r}{^{20}C_r}$ using the given identity $\frac{1}{^{2n+1}C_r} + \frac{1}{^{2n+1}C_{r+1}} = \frac{2n+2}{2n+1} \frac{1}{^{2n}C_r}$.

1. Apply the identity for the given sum: The sum involves $^{20}C_r$, so we set $2n=20$, which means $n=10$. Substitute $n=10$ into the identity: $\frac{1}{^{2(10)+1}C_r} + \frac{1}{^{2(10)+1}C_{r+1}} = \frac{2(10)+2}{2(10)+1} \frac{1}{^{20}C_r}$ $\frac{1}{^{21}C_r} + \frac{1}{^{21}C_{r+1}} = \frac{22}{21} \frac{1}{^{20}C_r}$ Rearranging this identity to express $\frac{1}{^{20}C_r}$: $\frac{1}{^{20}C_r} = \frac{21}{22} \left( \frac{1}{^{21}C_r} + \frac{1}{^{21}C_{r+1}} \right)$

2. Substitute into the summation: Let $S = \sum_{r=1}^{19} \frac{(-1)^r r}{^{20}C_r}$. Substitute the expression for $\frac{1}{^{20}C_r}$ into $S$: $S = \sum_{r=1}^{19} (-1)^r r \left[ \frac{21}{22} \left( \frac{1}{^{21}C_r} + \frac{1}{^{21}C_{r+1}} \right) \right]$ $S = \frac{21}{22} \sum_{r=1}^{19} \left( \frac{(-1)^r r}{^{21}C_r} + \frac{(-1)^r r}{^{21}C_{r+1}} \right)$ $S = \frac{21}{22} \left[ \sum_{r=1}^{19} \frac{(-1)^r r}{^{21}C_r} + \sum_{r=1}^{19} \frac{(-1)^r r}{^{21}C_{r+1}} \right]$

3. Rearrange the second sum: In the second sum, let $k = r+1$. Then $r = k-1$. When $r=1, k=2$. When $r=19, k=20$. $\sum_{r=1}^{19} \frac{(-1)^r r}{^{21}C_{r+1}} = \sum_{k=2}^{20} \frac{(-1)^{k-1} (k-1)}{^{21}C_k}$ So, $S = \frac{21}{22} \left[ \sum_{r=1}^{19} \frac{(-1)^r r}{^{21}C_r} + \sum_{r=2}^{20} \frac{(-1)^{r-1} (r-1)}{^{21}C_r} \right]$ (using $r$ instead of $k$ for consistency)

4. Combine the sums: The first term in the first sum is for $r=1$: $\frac{(-1)^1 \cdot 1}{^{21}C_1} = \frac{-1}{^{21}C_1}$. The last term in the second sum is for $r=20$: $\frac{(-1)^{20-1} (20-1)}{^{21}C_{20}} = \frac{(-1)^{19} \cdot 19}{^{21}C_{20}} = \frac{-19}{^{21}C_{20}}$. For $r$ from $2$ to $19$, we can combine the terms: $\frac{(-1)^r r}{^{21}C_r} + \frac{(-1)^{r-1} (r-1)}{^{21}C_r} = \frac{(-1)^{r-1} (-r + (r-1))}{^{21}C_r} = \frac{(-1)^{r-1} (-1)}{^{21}C_r} = \frac{(-1)^r}{^{21}C_r}$. So, $S = \frac{21}{22} \left[ \frac{-1}{^{21}C_1} + \sum_{r=2}^{19} \frac{(-1)^r}{^{21}C_r} + \frac{-19}{^{21}C_{20}} \right]$

5. Evaluate the sum $\sum_{r=2}^{19} \frac{(-1)^r}{^{21}C_r}$: Recall the property that for an odd integer $N$, $\sum_{k=0}^N \frac{(-1)^k}{^NC_k} = 0$. Here, $N=21$ (an odd integer). So, $\sum_{r=0}^{21} \frac{(-1)^r}{^{21}C_r} = 0$. Expand this sum: $\frac{(-1)^0}{^{21}C_0} + \frac{(-1)^1}{^{21}C_1} + \sum_{r=2}^{19} \frac{(-1)^r}{^{21}C_r} + \frac{(-1)^{20}}{^{21}C_{20}} + \frac{(-1)^{21}}{^{21}C_{21}} = 0$ We know $^{21}C_0 = 1$, $^{21}C_1 = 21$, $^{21}C_{20} = ^{21}C_{21-20} = ^{21}C_1 = 21$, $^{21}C_{21} = ^{21}C_{21-21} = ^{21}C_0 = 1$. Substitute these values: $1 - \frac{1}{21} + \sum_{r=2}^{19} \frac{(-1)^r}{^{21}C_r} + \frac{1}{21} - 1 = 0$ The terms $1$, $-\frac{1}{21}$, $\frac{1}{21}$, $-1$ cancel out. Therefore, $\sum_{r=2}^{19} \frac{(-1)^r}{^{21}C_r} = 0$.

6. Calculate the final value of S: Substitute this back into the expression for $S$: $S = \frac{21}{22} \left[ \frac{-1}{^{21}C_1} + 0 - \frac{19}{^{21}C_{20}} \right]$ $S = \frac{21}{22} \left[ \frac{-1}{21} - \frac{19}{21} \right]$ $S = \frac{21}{22} \left[ \frac{-1-19}{21} \right]$ $S = \frac{21}{22} \left[ \frac{-20}{21} \right]$ $S = -\frac{20}{22} = -\frac{10}{11}$

7. Determine p+q: The value is given as $-\frac{p}{q}$, where $p$ and $q$ are relatively prime natural numbers. Comparing $-\frac{10}{11}$ with $-\frac{p}{q}$, we get $p=10$ and $q=11$. 10 and 11 are relatively prime natural numbers. $p+q = 10+11 = 21$.

The final answer is $\boxed{21}$.

Explanation: The problem is solved by first using the given identity to express $\frac{1}{^{20}C_r}$ in terms of $^{21}C_r$ and $^{21}C_{r+1}$. This transforms the original sum into a new sum involving terms with $^{21}C_r$. By carefully splitting and combining terms in the new sum, it simplifies to a sum of the form $\sum \frac{(-1)^r}{^{21}C_r}$ and some boundary terms. The key property used is that for an odd integer $N$, the alternating sum of reciprocals of binomial coefficients $\sum_{k=0}^N \frac{(-1)^k}{^NC_k}$ is zero. This allows the central part of the sum to vanish, leaving only the boundary terms which are then easily calculated.