What is the value of sum S given by the following ?S=∑r=0n(−1)r( nCr r+3Cr)

nCr r+3Cr=n!r!(n−r)!r!3!(r+3)! =6.n!(n−r)!(r+3)!=6(n+1)(n+2)(n+3).(n+3)!(r+3)!(n−r)! S=6(n+1)(n+2)(n+3)∑r=0n(−1)r n+3Cr+3If we represents the summation expression in (1) by S1, then S1=∑r=0n(−1)r n+3Cr+3=∑s=3n+3(−1)s−3 n+3Cs=−∑s=3n+3(−1)s n+3Cs=−{∑s=0n+3(−1)s n+3Cs−(n+3C0−n+3C1+n+3C2}The term T1 is 0(why), while T2=1−(n+3)+(n+3)(n+2)2=(n+1)(n+2)2From (1) s=3n+3

What is the value of sum S given by the following ?S=∑r=0n(−1)r( nCr r+3Cr)

nCr r+3Cr=n!r!(n−r)!r!3!(r+3)! =6.n!(n−r)!(r+3)!=6(n+1)(n+2)(n+3).(n+3)!(r+3)!(n−r)! S=6(n+1)(n+2)(n+3)∑r=0n(−1)r n+3Cr+3If we represents the summation expression in (1) by S1, then S1=∑r=0n(−1)r n+3Cr+3=∑s=3n+3(−1)s−3 n+3Cs=−∑s=3n+3(−1)s n+3Cs=−{∑s=0n+3(−1)s n+3Cs−(n+3C0−n+3C1+n+3C2}The term T1 is 0(why), while T2=1−(n+3)+(n+3)(n+2)2=(n+1)(n+2)2From (1) s=3n+3

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What is the value of sum S given by the following ?
$S = \sum_{r = 0}^{n} {(- 1)}^{r} (\frac{^{n} C_{r}}{^{r + 3} C_{r}})$

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$\frac{3}{n + 3}$

$\frac{3}{n + 6}$

$\frac{6}{n + 3}$

$\frac{6}{n + 6}$

answer is A.

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Detailed Solution

$\frac{^{n} C_{r}}{^{r + 3} C_{r}} = \frac{n!}{r! (n - r)!} \frac{r! 3!}{(r + 3)!} = \frac{6. n!}{(n - r)! (r + 3)!} = \frac{6}{(n + 1) (n + 2) (n + 3)} . \frac{(n + 3)!}{(r + 3)! (n - r)!} S = \frac{6}{(n + 1) (n + 2) (n + 3)} \sum_{r = 0}^{n} {(- 1)}^{r}^{n + 3} C_{r + 3}$

If we represents the summation expression in (1) by $S_{1}$ , then
$S_{1} = \sum_{r = 0}^{n} {(- 1)}^{r}^{^{n + 3}} C_{r + 3} = \sum_{s = 3}^{n + 3} {(- 1)}^{s - 3 n + 3} C_{s} = - \sum_{s = 3}^{n + 3} {(- 1)}^{s n + 3} C_{s}$

$= - {\sum_{s = 0}^{n + 3} (- 1)^{s}^{n + 3} C_{s} - (^{n + 3} C_{0} -^{n + 3} C_{1} +^{n + 3} C_{2}}$
The term $T_{1}$ is 0(why), while
$T_{2} = 1 - (n + 3) + \frac{(n + 3) (n + 2)}{2} = \frac{(n + 1) (n + 2)}{2}$
From (1)